Most students don't like trigonometric inequalities. But in vain. As one character used to say,
“You just don’t know how to cook them”
So how to “cook” and with what to submit inequality with sine we will figure out in this article. We will decide in a simple way– using a unit circle.
So, first of all, we need the following algorithm.
Algorithm for solving inequalities with sine:
- on the sine axis we plot the number $a$ and draw a straight line parallel to the cosine axis until it intersects with the circle;
- the points of intersection of this line with the circle will be shaded if the inequality is not strict, and not shaded if the inequality is strict;
- the solution area of the inequality will be located above the line and up to the circle if the inequality contains the sign “$>$”, and below the line and up to the circle if the inequality contains the sign “$<$”;
- to find the intersection points, we solve the trigonometric equation $\sin(x)=a$, we get $x=(-1)^(n)\arcsin(a) + \pi n$;
- setting $n=0$, we find the first intersection point (it is located either in the first or fourth quarter);
- to find the second point, we look in which direction we go through the area to the second intersection point: if in a positive direction, then we should take $n=1$, and if in a negative direction, then $n=-1$;
- in response, the interval is written down from the smaller intersection point $+ 2\pi n$ to the larger one $+ 2\pi n$.
Algorithm limitation
Important: d given algorithm does not work for inequalities of the form $\sin(x) > 1; \ \sin(x) \geq 1, \ \sin(x)< -1, \ \sin{x} \leq -1$. В строгом случае эти неравенства не имеют решений, а в нестрогом – решение сводится к решению уравнения $\sin{x} = 1$ или $\sin{x} = -1$.
Special cases when solving inequalities with sine
It is also important to note following cases, which are much more convenient to solve logically without using the above algorithm.
Special case 1. Solve inequality:
$\sin(x)\leq 1.$
Due to the fact that the range of values of the trigonometric function $y=\sin(x)$ is not greater modulo $1$, then left side inequalities at any$x$ from the domain of definition (and the domain of definition of the sine is all real numbers) is not more than $1$. And, therefore, in the answer we write: $x \in R$.
Consequence:
$\sin(x)\geq -1.$
Special case 2. Solve inequality:
$\sin(x)< 1.$
Applying arguments similar to special case 1, we find that the left side of the inequality is less than $1$ for all $x \in R$, except for points that are solutions to the equation $\sin(x) = 1$. Solving this equation, we will have:
$x = (-1)^(n)\arcsin(1)+ \pi n = (-1)^(n)\frac(\pi)(2) + \pi n.$
And, therefore, in the answer we write: $x \in R \backslash \left\((-1)^(n)\frac(\pi)(2) + \pi n\right\)$.
Consequence: the inequality is solved similarly
$\sin(x) > -1.$
Examples of solving inequalities using an algorithm.
Example 1: Solve inequality:
$\sin(x) \geq \frac(1)(2).$
- Let us mark the coordinate $\frac(1)(2)$ on the sine axis.
- Let's draw a straight line parallel to the cosine axis and passing through this point.
- Let's mark the intersection points. They will be shaded because the inequality is not strict.
- The inequality sign is $\geq$, which means we paint the area above the line, i.e. smaller semicircle.
- We find the first intersection point. To do this, we turn the inequality into equality and solve it: $\sin(x)=\frac(1)(2) \ \Rightarrow \ x=(-1)^(n)\arcsin(\frac(1)(2) )+\pi n =(-1)^(n)\frac(\pi)(6) + \pi n$. We further set $n=0$ and find the first intersection point: $x_(1)=\frac(\pi)(6)$.
- We find the second point. Our area goes in the positive direction from the first point, which means we set $n$ equal to $1$: $x_(2)=(-1)^(1)\frac(\pi)(6) + \pi \cdot 1 = \ pi – \frac(\pi)(6) = \frac(5\pi)(6)$.
Thus, the solution will take the form:
$x \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right], \n \in Z.$
Example 2: Solve inequality:
$\sin(x)< -\frac{1}{2}$
Let's mark the coordinate $-\frac(1)(2)$ on the sine axis and draw a straight line parallel to the cosine axis and passing through this point. Let's mark the intersection points. They will not be shaded, since the inequality is strict. The inequality sign $<$, а, значит, закрашиваем область ниже прямой, т.е. меньший полукруг. Неравенство превращаем в равенство и решаем его:
$\sin(x)=-\frac(1)(2)$
$x=(-1)^(n)\arcsin(\left(-\frac(1)(2)\right))+ \pi n =(-1)^(n+1)\frac(\pi )(6) + \pi n$.
Further assuming $n=0$, we find the first intersection point: $x_(1)=-\frac(\pi)(6)$. Our area goes in the negative direction from the first point, which means we set $n$ equal to $-1$: $x_(2)=(-1)^(-1+1)\frac(\pi)(6) + \pi \cdot (-1) = -\pi + \frac(\pi)(6) = -\frac(5\pi)(6)$.
So, the solution to this inequality will be the interval:
$x \in \left(-\frac(5\pi)(6) + 2\pi n; -\frac(\pi)(6) + 2 \pi n\right), \n \in Z.$
Example 3: Solve inequality:
$1 – 2\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \leq 0.$
This example cannot be solved immediately using an algorithm. First you need to transform it. We do exactly what we would do with an equation, but don’t forget about the sign. Dividing or multiplying by a negative number reverses it!
So, let's move everything that does not contain a trigonometric function to the right side. We get:
$- 2\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \leq -1.$
Let's divide the left and right sides by $-2$ (don't forget about the sign!). Will have:
$\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \geq \frac(1)(2).$
Again we have an inequality that we cannot solve using an algorithm. But here it is enough to change the variable:
$t=\frac(x)(4)+\frac(\pi)(6).$
We obtain a trigonometric inequality that can be solved using the algorithm:
$\sin(t) \geq \frac(1)(2).$
This inequality was solved in Example 1, so let's borrow the answer from there:
$t \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right].$
However, the decision is not over yet. We need to go back to the original variable.
$(\frac(x)(4)+\frac(\pi)(6)) \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right].$
Let's imagine the interval as a system:
$\left\(\begin(array)(c) \frac(x)(4)+\frac(\pi)(6) \geq \frac(\pi)(6) + 2\pi n, \\ \frac(x)(4)+\frac(\pi)(6) \leq \frac(5\pi)(6) + 2 \pi n. \end(array) \right.$
On the left side of the system there is an expression ($\frac(x)(4)+\frac(\pi)(6)$), which belongs to the interval. The left boundary of the interval is responsible for the first inequality, and the right boundary is responsible for the second. Moreover, brackets play an important role: if the bracket is square, then the inequality will be relaxed, and if it is round, then it will be strict. our task is to get $x$ on the left in both inequalities.
Let's move $\frac(\pi)(6)$ from the left side to the right side, we get:
$\left\(\begin(array)(c) \frac(x)(4) \geq \frac(\pi)(6) + 2\pi n -\frac(\pi)(6), \\ \frac(x)(4) \leq \frac(5\pi)(6) + 2 \pi n – \frac(\pi)(6) \end(array) \right.$.
Simplifying, we have:
$\left\(\begin(array)(c) \frac(x)(4) \geq 2\pi n, \\ \frac(x)(4) \leq \frac(2\pi)(3) + 2 \pi n. \end(array) \right.$
Multiplying the left and right sides by $4$, we get:
$\left\(\begin(array)(c) x \geq 8\pi n, \\ x \leq \frac(8\pi)(3) + 8 \pi n. \end(array) \right. $
Assembling the system into the interval, we get the answer:
$x \in \left[ 8\pi n; \frac(8\pi)(3) + 8 \pi n\right], \n \in Z.$
When solving inequalities containing trigonometric functions, they are reduced to the simplest inequalities of the form cos(t)>a, sint(t)=a and similar ones. And already the simplest inequalities are solved. Let's look at various examples of ways to solve simple trigonometric inequalities.
Example 1. Solve the inequality sin(t) > = -1/2.
Draw a unit circle. Since sin(t) by definition is the y coordinate, we mark the point y = -1/2 on the Oy axis. We draw a straight line through it parallel to the Ox axis. At the intersection of the straight line with the graph of the unit circle, mark the points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by 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 it is necessary to indicate 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. To describe point Pt1, you can write the following formula:
t2 = pi - arcsin(-1/2) = 7*pi/6. As a result, we obtain the following inequality for t:
We preserve the inequalities. And since the sine function is periodic, it means that 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/6+2*pi*n< = t < = 7*pi/6 + 2*pi*n, при любом целом n.
Example 2. Solve cos(t) inequality<1/2.
Let's draw a unit circle. Since, according to the definition, cos(t) is the x coordinate, we mark the point x = 1/2 on the graph on the Ox axis.
We draw a straight line through this point parallel to the Oy axis. At the intersection of the straight line with the graph of the unit circle, mark the points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by two segments.
The solutions will be 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 the inequality for t: pi/3 Since 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 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. We will solve inequalities with tangent using the unit circle. Example 1: Solve inequality: $(\rm tg)(x) \leq 1.$ Thus, the solution will take the form: $x \in \left(-\frac(\pi)(2) + \pi n; \frac(\pi)(4) + \pi n\right], \ n \in Z.$ Important! Points $-\frac(\pi)(2)$ and $\frac(\pi)(2)$ at the tangent always (regardless of the inequality sign) gouged out! Example 2: Solve inequality: $(\rm tg)(x) > – \sqrt(3).$ We mark the point $- \sqrt(3)$ on the tangent line and draw a straight line from the origin to it. The point of intersection of this line with the semicircle will not be shaded, since the inequality is strict. The area will be located above the straight line and up to the circle, since the inequality sign is $>$. let's find the intersection point: $x_(1) = (\rm arctg)(\left(-\sqrt(3)\right)) = -\frac(\pi)(3).$ $t \in \left(-\frac(\pi)(3) + \pi n; \frac(\pi)(2) + \pi n\right).$ Let's return to the original variable: $\left(2x-\frac(\pi)(3)\right) \in \left(-\frac(\pi)(3) + \pi n; \frac(\pi)(2) + \pi n\right).$ The latter is equivalent to the system of inequalities $\left\(\begin(array)(c) 2x-\frac(\pi)(3) > -\frac(\pi)(3) + \pi n, \\ 2x-\frac(\pi) (3)< \frac{\pi}{2}+\pi n, \end{array} \right.$ having solved which we will get the answer. Really, $\left\(\begin(array)(c) 2x > \pi n, \\ 2x< \frac{5 \pi}{6} + \pi n, \end{array} \right.$ $\left\(\begin(array)(c) x > \frac(\pi n)(2), \\ x< \frac{5\pi}{12}+\frac{\pi n}{2}. \end{array} \right. $ And finally we get: $x \in \left(\frac(\pi n)(2); \frac(5\pi)(12) + \frac(\pi n)(2)\right), \n \in Z.$ 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 of coordinates, 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 --- periodic function, then to solve the inequality it is necessary to find its solutions 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 --- a bunch of all real numbers. Let 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. To find an interval that satisfies the conditions inequality sin x ‹ 1/2, you must perform the following steps: 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: Similar problems can be easily solved using a trigonometric circle. The algorithm for finding answers is very simple: 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. 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 already talking about 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 argumentAlgorithm for solving inequalities with tangent:
Examples of solving inequalities using an algorithm.
Method 1 - Solving inequalities by graphing a function
Method 2 - Solving trigonometric inequalities using the unit circle
Complex trigonometric inequalities
