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The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Trigonometric equations are all equations that include a variable under the sign of the trigonometric function. For example: \[\sin x= a, \cos x = b\]. Solving trigonometric equations comes down to the following subtasks:

* solving the equation;

* selection of roots.

The answer in such equations is written as:

degrees;

Radians.

To solve this kind of equation it is necessary to transform the equation into one/several basic trigonometric equations: \[\sin x = a; \cos x = a: \tan x = a; \cot x = a.\] And the solution to such basic equations is to use a conversion table or search for the positions of \[x\] on the unit circle.

For example, given trigonometric equations that can be solved using a conversion table of the following form:

\[\tan (x - \pi/4) = 0\]

Answer: \

\[\cot2x = 1.732\]

Answer: x = \[\pi /12 + \pi n\]

\[\sin x = 0.866\]

Answer: \[ x = \pi/3 \]

Where can I solve a system of trigonometric equations online for free?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

In this practical lesson Several typical examples will be considered that demonstrate methods for solving trigonometric equations and their systems.

This lesson will help you prepare for one of the types of tasks B5 and C1.

Preparation for the Unified State Exam in mathematics

Experiment

Lesson 10. Trigonometric functions. Trigonometric equations and their systems.

Practice

Lesson summary

We will devote the main part of the lesson to solving trigonometric equations and systems, but we will start with tasks on the properties of trigonometric functions that are not related to solving equations. Let's consider calculating the period of trigonometric functions with complex arguments.

Task No. 1. Calculate the period of functions a) ; b) .

Let's use the formulas given in the lecture.

a) For a function period . In our case, i.e. .

b) For the function period . With us, because the argument can be represented not only divided by three, but also multiplied by . Other operations with the function (multiplying by , adding 1) do not affect the argument, so we are not interested.

We get that

Answer. A) ; b) .

Let's move on to the main part of our practice and begin solving trigonometric equations. For convenience, we will analyze the solution to the same examples that we mentioned in the lecture when we listed the main types of equations.

Task No. 2. Solve the equation: a) ; b) ; V) ; G) .

To find the roots of such equations, we use formulas for general solutions.

To calculate the values ​​of the arc function, we use the oddity of the arc tangent and the table of values ​​of trigonometric functions, which we discussed in detail in the previous lesson. We will not dwell on these actions separately further.

d) When solving an equation, I would like to write using the general formula that , but this cannot be done. Here it is fundamentally important to check the range of cosine values, which is checked at the beginning of solving the equation.

Because the , which does not lie in the range of values ​​of the function, therefore, the equation has no solutions.

It is important not to confuse the value with the table value of cosine, be careful!

Comment. Quite often, in problems for solving trigonometric equations and systems, it is necessary to indicate not a general solution demonstrating an infinite family of roots, but to select only a few of them that lie in a certain range of values. Let's do these steps using the example of the answer to point “c”.

Additional task to point “c”. Indicate the number of roots of the equation that belong to the interval and list them.

We already know the general solution:

In order to indicate the roots belonging to the specified interval, they must be written out one by one, substituting specific parameter values. We will substitute integers starting from , because We are interested in roots from a range that is close to zero.

Upon substitution we get more higher value root, so there is no point in doing this. Now let's substitute negative values:

It makes no sense to substitute for the same reasons. Therefore, we have found the only root of the equation that belongs to the specified range.

Answer. ; the specified range contains one value of the root of the equation.

A similar formulation of the question of searching for certain values ​​of the roots of equations can also be found in tasks of other types; further we will not waste time on this. The search for the necessary roots will always be performed in the same way. Sometimes a trigonometric circle is depicted for this purpose. Try to plot on the circle the roots of the equations from points “a” and “b” that fall in the range.

Task No. 3. Solve the equation.

Let's use the method of finding roots using a trigonometric circle, as was shown in the lecture.

We draw points on the circle corresponding to the angles at which . There is only one such angle.

The first value of the angle corresponding to the specified point - the point is located on the ray, which is the origin. Next, in order to get to the same point again, but with a different angle value, you need to add to the first root found and get the next root . To obtain the next root, you must perform the same operation, etc.

Thus, we can indicate a general solution that will demonstrate that to obtain all the roots of the equation, it is necessary to add any integer number of times to the first value:

Let us recall that equations of the form can be solved in a similar way:

Task No. 4. Solve the equation .

The presence of a complex argument does not change the fact that the equation is, in fact, the simplest, and the approach to the solution remains the same. It’s just that now it acts as an argument. We write it in the general solution formula:

Problem #5. Solve the equation .

The most important thing is to prevent typical mistake and do not reduce both sides of the equation by , because in this case we will lose the roots of the equation corresponding to . A competent approach to solving involves moving all expressions to one side and adding a common factor.

At this stage, it is necessary to remember that if the product is equal to zero, then this is possible if either one of the factors is equal to zero or the other. Thus, our equation turns into a set of equations:

We solve the first equation as special case simplest equation. Do it yourself, we will write out the finished result. In the second equation, we will perform actions to bring it to its simplest form with a complex argument and solve it using the general formula of roots.

Please pay attention to this nuance - when recording general formula roots of the second equation we use another parameter "". This is due to the fact that we are solving a set of independent equations and they should not have common parameters. As a result, we obtain two independent families of solutions.

Answer. ; .

Problem #6. Solve the equation.

To simplify, we will use the formula for converting the product of trigonometric functions into the sum

Let's take advantage of the parity of the cosine and cancel out the same term in the two sides of the equation.

Let's move everything to one side and use the formula for the difference of cosines to get the product of functions, which will be equal to zero. Let's apply the formula for this .

Let's reduce both sides of the equation by:

We have reduced the equation to the product form that we got in the previous example. We suggest you work it out for yourself. Let's indicate the final answer.

In principle, this is the final answer. However, it can be written more compactly as one family of solutions rather than two. The first solution contains all the quarters of the parts, and the second contains all the halves of the parts, but the halves are included in the quarters, since half is two quarters. Thus, the second family of roots is included in the first, and the final answer can be described by the first family of solutions.

To better understand these arguments, try plotting the resulting roots on a trigonometric circle.

Answer. or .

We looked at one equation using transformations of trigonometric functions, but there are a huge number of them, as well as types of transformations. We will consider the equation for using universal trigonometric substitution, an example of which we did not give in the lesson before last, after we analyze the substitution method.

Problem No. 7. Solve the equation.

In this case, you must first try to reduce the equation to using a single trigonometric function. Because is easily expressed through using the trigonometric unit, we can easily reduce the equation to sines.

Let's substitute the expression into our equation:

Since everything is reduced to one function, we can perform the replacement: .

We have obtained a quadratic equation that can be easily solved in any way convenient for you, for example, using Vieta’s theorem it is easy to obtain that:

The first equation has no solutions, because the sine value is outside the permissible region.

We suggest you solve the second equation yourself, because... This is the type of special cases of the simplest equations that we have already considered. Let's write down its roots:

Answer. .

Problem No. 8. Solve the equation.

In this equation, the solution methods that we have already considered are not immediately visible. In such cases, you should try to apply the formulas of universal trigonometric substitution, which will help reduce the equation to a single function.

Let's use the formulas: and , which will bring the entire equation to .

Now it is clear that it is possible to perform a replacement.

Let's add the fractions and multiply both sides of the equation by the denominator, because it is not equal to zero.

We have reduced the equation to the form already discussed earlier, i.e. to the product of factors, which is equal to zero.

Let's do the reverse substitution:

Both resulting families of solutions can be easily combined into one:

Answer. .

Problem No. 9. Solve the equation. In your answer, provide only roots that are multiples of .

The indicated equation becomes more complicated after reduction to sines or cosines, as one would like to do using the trigonometric unit formula. Therefore, another method is used.

We called the indicated equation homogeneous, this is the name given to equations in which, after rearranging unknown functions or variables, nothing will change. Swap the sine and cosine and you will see that this is our case.

Homogeneous equations are solved by dividing both sides by the highest power of the function. In our case it is either or. We choose the one we like best and divide both sides of the equation by it. Let's take for example this . In this case, it is imperative to check whether during such a division we will not lose the roots corresponding to , i.e. . To do this, first substitute into the original equation.

Since we did not obtain an identity, the roots of our equation will not correspond.

Now we can safely divide by:

We have reduced the equation to substitution, and this method of solution has already been considered. As they say, “let’s pour the water out of the kettle” and reduce the problem to what is already known. Decide further yourself. We will indicate the final answer:

Since in the problem statement we are required to indicate only multiple roots, we will write down only the first family of solutions as an answer.

Problem No. 10. Solve the equation .

This equation is surprising in that it contains two unknowns, and as we know, in the general case such an equation cannot be solved. Another problem is that this equation is fundamentally different from all those discussed previously, because the unknown in it is not only in the argument of the trigonometric function.

To solve it, let's pay attention to the properties of functions that are equal on the left and on the right. Specifically, we are interested in what values ​​these functions are limited to.

For cosine we know the range of values:

For a quadratic function:

From this we can conclude that these expressions can only have one general meaning, when each of them is equal to 1. We obtain a system of equations:

Both equations turn out to be independent and contain one variable each, so they can be easily solved using methods already known to us.

Of course specified method is not obvious, and the task is one of increased complexity. This method is sometimes called “mini-max”, because the equality of the minimum and maximum value functions.

Now we will consider separately methods for solving systems of trigonometric equations. The methods for solving them are standard, we will just use formulas for transformations of trigonometric functions. Let's look at the most common types of such systems.

Problem No. 11. Solve system of equations .

We solve by substitution method, express from a simpler linear equation, for example, and substitute it into the second equation:

In the second equation we use what is the period of the sine, i.e. it can be removed, and sine is an odd function, i.e. a minus is taken out of it.

According to the addition formula harmonic vibrations we reduce the second equation to one trigonometric function. Try these conversions yourself.

Let us substitute the resulting solution into the expression for:

In this case, we use the same parameter for both families of solutions, because they are dependent on each other.

Systems of simple trigonometric equations.

Problem No. 12. Solve system of equations .

Both equations in the system are special cases of the simplest equations, we know how to solve them, and the system quickly reduces to linear.

The parameters in both equations are different, because we solved the equations independently of each other and the variables were not yet expressed one through the other.

Now let's decide linear system using the substitution or addition method, as you prefer, do these steps yourself. Let's indicate the final result.

Pay attention to the recording of the solution of the system when the variables depend simultaneously on two parameters. In order to write down the numerical values ​​of the roots, in this case, all integer values ​​of the parameters that do not depend on each other are substituted in turn.

In this practical part of the lesson, we looked at several typical examples in which we demonstrated methods for solving trigonometric equations and their systems.

Lessons 54-55. Systems of trigonometric equations (optional)

09.07.2015 9315 915

Target: consider the most typical systems of trigonometric equations and methods for solving them.

I. Communicating the topic and purpose of the lessons

II. Repetition and consolidation of the material covered

1. Answers to questions about homework(analysis of unsolved problems).

2. Monitoring the assimilation of the material (independent work).

Option 1

Solve the inequality:

Option 2

Solve the inequality:

III. Learning new material

In exams, systems of trigonometric equations are much less common than trigonometric equations and inequalities. There is no clear classification of systems of trigonometric equations. Therefore, we will conditionally divide them into groups and consider ways to solve these problems.

1. The simplest systems of equations

These include systems in which either one of the equations is linear, or the equations of the system can be solved independently of each other.

Example 1

Let's solve the system of equations

Since the first equation is linear, we express the variable from itand substitute into the second equation:We use the reduction formula and the main trigonometric identity. We get the equation or Let's introduce a new variable t = sin u. We have quadratic equation 3 t 2 - 7 t + 2 = 0, whose roots t 1 = 1/3 and t 2 = 2 (not suitable because sin y ≤ 1). Let's return to the old unknown and get the equation siny = 1/3, whose solutionNow it's easy to find the unknown:So, the system of equations has solutions where n ∈ Z.

Example 2

Let's solve the system of equations

The equations of the system are independent. Therefore, we can write down the solutions to each equation. We get:We add and subtract the equations of this system of linear equations term by term and find:where

Please note that due to the independence of the equations, when finding x - y and x + y, different integers must be specified n and k. If instead of k was also supplied n , then the solutions would look like:In this case, an infinite number of solutions would be lost and, in addition, a connection between the variables would arise x and y: x = 3y (which is not the case in reality). For example, it is easy to check that this system has a solution x = 5π and y = n (in accordance with the formulas obtained), which when k = n impossible to find. So be careful.

2. Type systems

Such systems are reduced to the simplest by adding and subtracting equations. In this case we obtain systemsor Let's note an obvious limitation: And The solution of such systems itself does not present any difficulties.

Example 3

Let's solve the system of equations

Let us first transform the second equation of the system using the equality We get: Let's substitute the first equation into the numerator of this fraction:and express Now we have a system of equationsLet's add and subtract these equations. We have: orLet us write down the solutions to this simplest system:Adding and subtracting these linear equations, we find:

3. Type systems

Such systems can be considered as simplest and solved accordingly. However, there is another way to solve it: convert the sum of the trigonometric functions into a product and use the remaining equation.

Example 4

Let's solve the system of equations

First, we transform the first equation using the formula for the sum of the sines of angles. We get:Using the second equation, we have:where Let us write down the solutions to this equation:Taking into account the second equation of this system, we obtain a system of linear equationsFrom this system we find It is convenient to write such decisions in a more rational form. For the upper signs we have:for lower signs -

4. Type systems

First of all, it is necessary to obtain an equation containing only one unknown. To do this, for example, let us express from one equation sin y, from another - cos u. Let's square these ratios and add them up. Then we get a trigonometric equation containing the unknown x. Let's solve this equation. Then, using any equation of this system, we obtain an equation for finding the unknown y.

Example 5

Let's solve the system of equations

Let us write the system in the formLet us square each equation of the system and get:Let's add up the equations of this system: or Using the basic trigonometric identity, we write the equation in the form or Solutions to this equation cos x = 1/2 (then ) and cos x = 1/4 (from where ), where n, k ∈ Z . Considering the connection between the unknowns cos y = 1 – 3 cos x, we get: for cos x = 1/2 cos y = -1/2; for cos x = 1/4 cos y = 1/4. It must be remembered that when solving a system of equations, squaring was carried out and this operation could lead to the appearance of extraneous roots. Therefore, it is necessary to take into account the first equation of this system, from which it follows that the quantities sin x and sin y must have the same sign.

Taking this into account, we obtain solutions to this system of equationsAnd where n, m, k, l ∈ Z . In this case, for unknown x and y, either upper or lower signs are simultaneously chosen.

In a special casethe system can be solved by converting the sum (or difference) of trigonometric functions into a product and then dividing the equations term by term.

Example 6

Let's solve the system of equations

In each equation, we transform the sum and difference of the functions into a product and divide each equation by 2. We get:Since not a single factor on the left sides of the equations is equal to zero, we divide the equations term by term (for example, the second by the first). We get:where Let's substitute the found valuefor example, in the first equation:Let's take into account that Then where

We obtained a system of linear equationsBy adding and subtracting the equations of this system, we findAnd where n, k ∈ Z.

5. Systems solved by replacing unknowns

If the system contains only two trigonometric functions or can be reduced to this form, then it is convenient to use the replacement of unknowns.

Example 7

Let's solve the system of equations

Since this system includes only two trigonometric functions, we introduce new variables a = tan x and b = sin u. We obtain a system of algebraic equationsFrom the first equation we express a = b + 3 and substitute into the second:or The roots of this quadratic equation b 1 = 1 and b 2 = -4. The corresponding values ​​are a1 = 4 and a2 = -1. Let's return to the old unknowns. We obtain two systems of simple trigonometric equations:

a) her decision where n, k ∈ Z.

b) has no solutions, because sin y ≥ -1.

Example 8

Let's solve the system of equations

Let us transform the second equation of the system so that it contains only the functions sin x and cos u. To do this, we use the reduction formulas. We get:(where ) And (Then ). The second equation of the system has the form: or We obtained a system of trigonometric equationsLet's introduce new variables a = sin x and b = cos u. We have a symmetric system of equations the only solution to which a = b = 1/2. Let's return to the old unknowns and get the simplest system of trigonometric equations the solution of which where n, k ∈ Z.

6. Systems for which the features of the equations are important

Almost when solving any system of equations, one or another of its features is used. In particular, one of the most general techniques solutions of the system are identical transformations that make it possible to obtain an equation containing only one unknown. The choice of transformations, of course, is determined by the specifics of the system equations.

Example 9

Let's solve the system

Let us pay attention to the left-hand sides of the equations, for example toUsing reduction formulas, we make it a function with argument π/4 + x. We get:Then the system of equations looks like:To eliminate the variable x, we multiply the equations term by term and get:or 1 = sin 3 2у, whence sin 2у = 1. We find And It is convenient to consider separately the cases of even and odd values n. For even n (n = 2 k, where k ∈ Z) Then from the first equation of this system we obtain:where m ∈ Z. For odd Then from the first equation we have:So, this system has solutions

As in the case of equations, there are quite often systems of equations in which the limited nature of the sine and cosine functions plays a significant role.

Example 10

Let's solve the system of equations

First of all, we transform the first equation of the system:or or or or Taking into account the limited nature of the sine function, we see that the left side of the equation is not less than 2, and right part no more than 2. Therefore, such an equation is equivalent to the conditions sin 2 2x = 1 and sin 2 y = 1.

We write the second equation of the system in the form sin 2 y = 1 - cos 2 z or sin 2 y = sin 2 z, and then sin 2 z = 1. We obtained a system of simple trigonometric equationsUsing the formula for reducing the degree, we write the system in the formor Then

Of course, when solving other systems of trigonometric equations, it is also necessary to pay attention to the features of these equations.

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When solving many mathematical problems , especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic ones. Principle successful solution each of the mentioned tasks is as follows: it is necessary to establish what type of problem is being solved, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to have the skills to perform identity transformations and computing.

The situation is different with trigonometric equations. It is not at all difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer.

By appearance equation, it is sometimes difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.

To solve a trigonometric equation, you need to try:

1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. unfold left side factoring equations, etc.

Let's consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution diagram

Step 1. Express a trigonometric function in terms of known components.

Step 2. Find the function argument using the formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

sin x = a; x = (-1) n arcsin a + πn, n Є Z.

tan x = a; x = arctan a + πn, n Є Z.

ctg x = a; x = arcctg a + πn, n Є Z.

Step 3. Find the unknown variable.

Example.

2 cos(3x – π/4) = -√2.

Solution.

1) cos(3x – π/4) = -√2/2.

2) 3x – π/4 = ±(π – π/4) + 2πn, n Є Z;

3x – π/4 = ±3π/4 + 2πn, n Є Z.

3) 3x = ±3π/4 + π/4 + 2πn, n Є Z;

x = ±3π/12 + π/12 + 2πn/3, n Є Z;

x = ±π/4 + π/12 + 2πn/3, n Є Z.

Answer: ±π/4 + π/12 + 2πn/3, n Є Z.

II. Variable replacement

Solution diagram

Step 1. Reduce the equation to algebraic form with respect to one of the trigonometric functions.

Step 2. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).

Step 3. Write down and solve the resulting algebraic equation.

Step 4. Make a reverse replacement.

Step 5. Solve the simplest trigonometric equation.

Example.

2cos 2 (x/2) – 5sin (x/2) – 5 = 0.

Solution.

1) 2(1 – sin 2 (x/2)) – 5sin (x/2) – 5 = 0;

2sin 2 (x/2) + 5sin (x/2) + 3 = 0.

2) Let sin (x/2) = t, where |t| ≤ 1.

3) 2t 2 + 5t + 3 = 0;

t = 1 or e = -3/2, does not satisfy the condition |t| ≤ 1.

4) sin(x/2) = 1.

5) x/2 = π/2 + 2πn, n Є Z;

x = π + 4πn, n Є Z.

Answer: x = π + 4πn, n Є Z.

III. Equation order reduction method

Solution diagram

Step 1. Replace this equation with a linear one, using the formula for reducing the degree:

sin 2 x = 1/2 · (1 – cos 2x);

cos 2 x = 1/2 · (1 + cos 2x);

tg 2 x = (1 – cos 2x) / (1 + cos 2x).

Step 2. Solve the resulting equation using methods I and II.

Example.

cos 2x + cos 2 x = 5/4.

Solution.

1) cos 2x + 1/2 · (1 + cos 2x) = 5/4.

2) cos 2x + 1/2 + 1/2 · cos 2x = 5/4;

3/2 cos 2x = 3/4;

2x = ±π/3 + 2πn, n Є Z;

x = ±π/6 + πn, n Є Z.

Answer: x = ±π/6 + πn, n Є Z.

IV. Homogeneous equations

Solution diagram

Step 1. Reduce this equation to the form

a) a sin x + b cos x = 0 ( homogeneous equation first degree)

or to the view

b) a sin 2 x + b sin x · cos x + c cos 2 x = 0 (homogeneous equation of the second degree).

Step 2. Divide both sides of the equation by

a) cos x ≠ 0;

b) cos 2 x ≠ 0;

and get the equation for tan x:

a) a tan x + b = 0;

b) a tan 2 x + b arctan x + c = 0.

Step 3. Solve the equation using known methods.

Example.

5sin 2 x + 3sin x cos x – 4 = 0.

Solution.

1) 5sin 2 x + 3sin x · cos x – 4(sin 2 x + cos 2 x) = 0;

5sin 2 x + 3sin x · cos x – 4sin² x – 4cos 2 x = 0;

sin 2 x + 3sin x · cos x – 4cos 2 x = 0/cos 2 x ≠ 0.

2) tg 2 x + 3tg x – 4 = 0.

3) Let tg x = t, then

t 2 + 3t – 4 = 0;

t = 1 or t = -4, which means

tg x = 1 or tg x = -4.

From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.

Answer: x = π/4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.

V. Method of transforming an equation using trigonometric formulas

Solution diagram

Step 1. Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.

Step 2. Solve the resulting equation using known methods.

Example.

sin x + sin 2x + sin 3x = 0.

Solution.

1) (sin x + sin 3x) + sin 2x = 0;

2sin 2x cos x + sin 2x = 0.

2) sin 2x (2cos x + 1) = 0;

sin 2x = 0 or 2cos x + 1 = 0;

From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.

We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.

As a result, x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

Answer: x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

The ability and skill to solve trigonometric equations is very important, their development requires significant effort, both on the part of the student and on the part of the teacher.

Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems embodies many of the knowledge and skills that are acquired by studying the elements of trigonometry.

Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.

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