Example 3. Solve inequality tg(t)< = 1.
The tangent period is equal to pi. Let's find solutions that belong to the interval (-pi/2;pi/2) right semicircle. Next, using the periodicity of the tangent, we write down all the solutions to this inequality. Let's draw a unit circle and mark a 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, point P(-pi/2) does not belong to this arc.
Solving trigonometric inequalities using the unit circle
When solving trigonometric inequalities of the form, where --- one of the trigonometric functions, it is convenient to use the trigonometric circle in order to most clearly represent the solutions to the inequality and write down the answer. The main method for solving trigonometric inequalities is to reduce them to the simplest type inequalities. Let's look at an example of how to solve such inequalities.
Example Solve the inequality.
Solution. Let's draw a trigonometric circle and mark on it the points for which the ordinate is superior.
To solve this inequality there will be. It is also clear that if a certain number differs from any number from the specified interval by, then it will also be no less. Therefore, you just need to add solutions to the ends of the found segment. Finally, we find that all solutions to the original inequality will be.
To solve inequalities with tangent and cotangent, the concept of a line of tangents and cotangents is useful. These are the straight lines and, respectively (in Figure (1) and (2)), tangent to the trigonometric circle.
It is easy to see that if we construct a ray with its origin at the origin, making an angle with the positive direction of the abscissa axis, then the length of the segment from the point to the point of intersection of this ray with the tangent line is exactly equal to the tangent of the angle that this ray makes with the abscissa axis. A similar observation occurs for cotangent.
Example Solve the inequality.
Solution. Let us denote, then the inequality will take the simplest form: . Let's consider an interval of length equal to the smallest positive period (LPP) of the tangent. On this segment, using the line of tangents, we establish that. Let us now remember what needs to be added since NPP functions. So, . Returning to the variable, we get that
It is convenient to solve inequalities with inverse trigonometric functions using graphs of inverse trigonometric functions. Let's show how this is done with an example.
Solving trigonometric inequalities graphically
Note that if --- a periodic function, then to solve the inequality it is necessary to find its solution on a segment whose length is equal to the period of the function. All solutions to the original inequality will consist of the found values, as well as all those that differ from those found by any integer number of periods of the function
Let's consider the solution to inequality ().
Since, then the inequality has no solutions. If, then the set of solutions to the inequality is the set of all real numbers.
Let it be. The sine function has the smallest positive period, so the inequality can be solved first on a segment of length, e.g. We build graphs of functions and ().
On the segment, the sine function increases, and the equation, where, has one root. On the segment, the sine function decreases, and the equation has a root. On a numerical interval, the graph of a function is located above the graph of the function. Therefore, for all from the interval) the inequality holds if. Due to the periodicity of the sine function, all solutions to the inequality are given by inequalities of the form: .
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 graphing a function
To find an interval that satisfies the conditions of inequality sin x ‹ 1/2, you must perform the following steps:
- On the coordinate axis, construct a sinusoid y = sin x.
- On the same axis, draw a graph of the numerical argument of the inequality, i.e., a straight line passing through the point ½ of the ordinate OY.
- Mark the intersection points of the two graphs.
- Shade the segment that is the solution to the example.
When strict signs are present in an expression, the intersection points are not solutions. Since the smallest positive period of a 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 the following inequality:
Method 2 - Solving trigonometric inequalities using the unit circle
Similar problems can be easily solved using a trigonometric circle. The algorithm for finding answers is very simple:
- First you need to draw a unit circle.
- 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 a circle.
- It is necessary to draw a straight line passing through the value of the arc function parallel to the abscissa axis (OX).
- After that, all that remains is to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
- Write down the answer in the required form.
Let us analyze the stages of the solution using the example of the inequality sin x › 1/2. Points α and β are marked on the circle - 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 answer arc will be located symmetrically to the OX axis, 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.
Arctangent and arccotangent are tangents to a trigonometric circle, and the minimum positive period for both functions is π. To quickly and correctly use the second method, you need to remember on which axis the values of sin, cos, tg and ctg are plotted.
The tangent tangent runs parallel to the OY axis. If we plot the value of arctan a on the unit circle, then the second required point will be located in the diagonal quarter. Angles
They are break points for the function, since the graph tends to them, but never reaches them.
In the case of cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at 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 process and procedure for solving it are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:
The graphical solution involves constructing an ordinary sinusoid y = sin x using arbitrarily selected values of x. Let's calculate a table with coordinates for the control points of the graph:
The result should be a beautiful curve.
To make finding a solution easier, let’s replace the complex function argument