Inequalities are relations of the form a › b, where a and b are expressions containing at least one variable. Inequalities can be strict - ‹, › and non-strict - ≥, ≤.

Trigonometric inequalities are expressions of the form: F(x) › a, F(x) ‹ a, F(x) ≤ a, F(x) ≥ a, in which F(x) is represented by one or more trigonometric functions.

An example of the simplest trigonometric inequality is: sin x ‹ 1/2. It is customary to solve such problems graphically; two methods have been developed for this.

Method 1 - Solving Inequalities by Plotting a Function

To find an interval that satisfies the inequality sin x ‹ 1/2, you must do the following:

1. On the coordinate axis, construct a sinusoid y = sin x.
2. On the same axis, draw a graph of the numerical argument of the inequality, that is, a straight line passing through the point ½ of the OY ordinate.
3. Mark the intersection points of the two graphs.
4. Shade the segment that is the solution of the example.

When there are strong signs in an expression, the intersection points are not solutions. Since the smallest positive period of the sinusoid is 2π, we write the answer as follows:

If the signs of the expression are not strict, then the solution interval must be enclosed in square brackets - . The answer to the problem can also be written as another inequality:

Method 2 - Solving trigonometric inequalities using the unit circle

Similar problems are easily solved with the help of a trigonometric circle. The search algorithm is very simple:

1. First, draw a unit circle.
2. Then you need to note the value of the arc function of the argument of the right side of the inequality on the arc of the circle.
3. It is necessary to draw a straight line passing through the value of the arc function parallel to the x-axis (OX).
4. After that, it remains only to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
5. Write the answer in the required form.

Let us analyze the solution steps using the inequality sin x › 1/2 as an example. Points α and β are marked on the circle – the values

The points of the arc located above α and β are the interval for solving the given inequality.

If you need to solve an example for cos, then the arc of answers will be located symmetrically to the OX axis, and not OY. You can consider the difference between the solution intervals for sin and cos in the diagrams below in the text.

Graphical solutions for tangent and cotangent inequalities will differ from both sine and cosine. This is due to the properties of functions.

The arctangent and arccotangent are tangents to the trigonometric circle, and the minimum positive period for both functions is π. In order to quickly and correctly use the second method, you need to remember on which axis the values ​​\u200b\u200bof sin, cos, tg and ctg are plotted.

The tangent tangent runs parallel to the OY axis. If we plot the value of arctg a on the unit circle, then the second required point will be located in the diagonal quarter. corners

They are breakpoints for the function, as the graph tends to them but never reaches them.

In the case of the cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at the points π and 2π.

Complex trigonometric inequalities

If the argument of the inequality function is represented not just by a variable, but by an entire expression containing an unknown, then we are talking about a complex inequality. The course and order of its solution are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:

The graphical solution provides for the construction of an ordinary sinusoid y = sin x for arbitrarily chosen values ​​of x. Let's calculate a table with coordinates for the chart's reference points:

The result should be a nice curve.

For ease of finding a solution, we replace the complex function argument

The intersection of two graphs allows you to determine the area of ​​the desired values ​​for which the inequality condition is satisfied.

The found segment is the solution for the variable t:

However, the goal of the task is to find all possible variants of the unknown x:

Solving the double inequality is quite simple, you need to move π / 3 to the extreme parts of the equation and perform the required calculations:

Answer to the task will look like an interval for strict inequality:

Such tasks will require the experience and skill of students in handling trigonometric functions. The more training tasks will be solved in the process of preparation, the easier and faster the student will find the answer to the question of the exam test.

Inequalities containing trigonometric functions, when solved, are reduced to the simplest inequalities of the form cos(t)>a, sint(t)=a and the like. And already the simplest inequalities are solved. Consider on various examples ways to solve the simplest trigonometric inequalities.

Example 1. Solve the inequality sin(t) > = -1/2.

Draw a single circle. Since sin (t) by definition is the y coordinate, we mark the point y \u003d -1/2 on the Oy axis. We draw a straight line through it parallel to the x-axis. At the points of intersection of the straight line with the graph of the unit circle, we mark the points Pt1 and Pt2. We connect the origin of coordinates with the points Pt1 and Pt2 with two segments.

The solution to this inequality will be all points of the unit circle located above these points. In other words, the solution will be the arc l.. Now you need to specify the conditions under which an arbitrary point will belong to the arc l.

Pt1 lies in the right semicircle, its ordinate is -1/2, then t1=arcsin(-1/2) = - pi/6. The following formula can be written to describe the point Pt1:
t2 = pi - arcsin(-1/2) = 7*pi/6. As a result, we obtain the following inequality for t:

We keep the inequality signs. And since the sine function is a periodic function, then the solutions will be repeated every 2 * pi. We add this condition to the obtained inequality for t and write down the answer.

Answer: -pi/6+2*pi*n< = t < = 7*pi/6 + 2*pi*n, при любом целом n.

Example 2 Solve the inequality cos(t)<1/2.

Let's draw a unit circle. Since, according to the definition of cos(t), this is the x-coordinate, we mark the point x = 1/2 on the graph on the x-axis.
We draw a straight line through this point parallel to the y-axis. At the points of intersection of the straight line with the graph of the unit circle, we mark the points Pt1 and Pt2. We connect the origin of coordinates with the points Pt1 and Pt2 with two segments.

The solutions are all points of the unit circle that belong to the arc l.. Let's find the points t1 and t2.

t1 = arccos(1/2) = pi/3.

t2 = 2*pi - arccos(1/2) = 2*pi-pi/3 = 5*pi/6.

We got an inequality for t: pi/3

Since the cosine is a periodic function, the solutions will be repeated every 2 * pi. We add this condition to the resulting inequality for t and write down the answer.

Answer: pi/3+2*pi*n

Example 3 Solve the inequality tg(t)< = 1.

The period of the tangent is pi. Let's find solutions that belong to the interval (-pi/2;pi/2) the right semicircle. Next, using the periodicity of the tangent, we write down all solutions of this inequality. Let's draw a unit circle and mark the line of tangents on it.

If t is a solution to the inequality, then the ordinate of the point T = tg(t) must be less than or equal to 1. The set of such points will make up the ray AT. The set of points Pt that will correspond to the points of this ray is the arc l. Moreover, the point P(-pi/2) does not belong to this arc.

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Solving inequalities containing trigonometric functions usually comes down to solving the simplest inequalities of the form: sin(t);≥)a;cos(t);≥)a;tg(t);≥)a;ctg(t);≥)a; The ways of solving these inequalities follow quite obviously from the representation of trigonometric functions on the unit circle.

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Inequalities: sin x > a, sin x a, sin x

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Trigonometric inequality sin(t)≥a.

All points Pt of the unit circle for values ​​of t satisfying this inequality have an ordinate greater than or equal to -1/2. The set of such points is the arc l, which is highlighted in bold in the figure below. Let us find the condition for the point Pt to belong to this arc. The point Pt lies on the right semicircle, the ordinate of Pt is equal to 1/2, and, therefore, it is convenient to take the value t1=arcsin(-1/2)=-π/6 as t1. Imagine that we are going around the arc l from the point Pt1 to Pt2 counterclockwise. Then t2 > t1, and, as is easy to understand, t2=π-arcsin(-1/2)=7*π/6. Thus, we get that the point Pt belongs to the arc l if -π/6 ≤ t ≤ 7*π/6. Thus, the solutions of the inequality belonging to the interval [-π/2 ; 3*π/2] of length 2*π are: -π/6 ≤ t ≤ 7*π/6. Due to the periodicity of the sine, the remaining solutions are obtained by adding to the found numbers of the form 2πn, where n is an integer. Thus, we arrive at the answer: -π/6+2πn≤t≤7π/6+2πn, n is an integer.

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Example 1

Solve the inequality Let's draw a trigonometric circle and mark on it the points for which the ordinate exceeds For x, the solution to this inequality will be It is also clear that if some number x differs from any number from the specified interval by 2π n then sin x will also be no less Therefore , to the ends of the found segment of the solution, you just need to add 2π n , where Finally, we get that the solutions of the original inequality will be all where Answer. where

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Trigonometric inequality cos(t)

Consider the solution of the simplest trigonometric inequalities with cosine using the example of solving the inequality cos(t) t1 and t2=2π-arccos(1/2)=5π/3. The point belongs to the distinguished arc l (excluding its ends) provided that π/3