Precalculus

Trigonometry

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Some Other Trigonometric Inequalities

A trigonometric inequality is called an inequality that contains a trigonometric function (sin x, cos x, tan x, cot x) of an unknown angle.

Solving trigonometric inequalities (as well as equations), as a rule, can be reduced to solving the basic trigonometric inequalities such as

\[\sin x \gt a, \cos x \le a, \tan x \lt a, \cot \ge a.\]

The simplest trigonometric inequalities are conveniently solved using the unit circle.

If the left side of a trigonometric inequality \(f\left({x}\right) \ge 0\) (the inequality sign can also be \({\ge,\lt \text{ or } \le}\)) is represented as

\[f\left({x}\right) = \left({f\left({x}\right) - a_1}\right) \left({f\left({x}\right) - a_2}\right) \cdots \left({f\left({x}\right) - a_n}\right),\]

where \(f\left({x}\right)\) is one of the trigonometric functions \(\sin x,\) \(\cos x,\) \(\tan x,\) \(\cot x\) and \({a_1, a_2, \ldots, a_n}\) are real numbers, then such inequality can be solved by the interval method.

Using the method of intervals on a unit circle, it should be remembered that if at some point there is an odd number of coinciding roots, then the sign of the expression

\[f\left({x}\right) = \left({f\left({x}\right) - a_1}\right) \left({f\left({x}\right) - a_2}\right) \cdots \left({f\left({x}\right) - a_n}\right)\]

changes to the opposite when passing through such a point. If there is an even number of roots at the point, then the sign of the expression does not change.

Solved Problems

Example 1.

Solve the inequality

\[\tan x + \cot x \le 2.\]

Solution.

Using the definition of the cotangent, we write the inequality in the following form:

\[\tan x + \frac{1}{\tan x} - 2 \le 0, \Rightarrow \tan^2x - 2\tan x + 1 \le 0.\]

We are assuming here that the tangent is not zero, which is obvious.

The last inequality is written as

\[\left({\tan x - 1}\right)^2 \le 0.\]

Its solution is possible only in the form of equality

\[\tan x - 1 = 0.\]

Hence,

\[\tan x = 1, \Rightarrow x = \frac{\pi}{4} + \pi n,\;n \in \mathbb{Z}.\]

Example 2.

Solve the inequality

\[\sin\left({3x - \frac{\pi}{6}}\right)\sin\left({3x + \frac{\pi}{6}}\right) \ge 0.\]

Solution.

We convert the product of functions to a sum:

\[\sin\left({3x - \frac{\pi}{6}}\right)\sin\left({3x + \frac{\pi}{6}}\right) = \frac{1}{2}\left[{\cos\left({\cancel{3x}-\frac{\pi}{6}-\cancel{3x}-\frac{\pi}{6}}\right) - \cos\left({3x-\cancel{\frac{\pi}{6}}+3x+\cancel{\frac{\pi}{6}}}\right)}\right] = \frac{1}{2}\left[{\cos\left({-\frac{\pi}{3}}\right) - \cos6x}\right].\]

Since

\[\cos\left({-\frac{\pi}{3}}\right) = \cos\frac{\pi}{3} = \frac{1}{2},\]

the inequality takes the form

\[\frac{1}{2}\left({\frac{1}{2} - \cos6x}\right) \ge 0, \Rightarrow \cos6x \le \frac{1}{2}.\]

We have obtained the basic inequality. Its solution looks like

\[\frac{\pi}{3} + 2\pi n \le 6x \le \frac{5\pi}{3} + 2\pi n,\]
\[\Rightarrow \frac{\pi}{18} + \frac{\pi n}{3} \le x \le \frac{5\pi}{18} + \frac{\pi n}{3},\;n \in \mathbb{Z}.\]

Example 3.

Solve the inequality

\[\frac{2}{\sin x \cos x} \gt 0.\]

Solution.

Using the double-angle identity for sine, we write the inequality in the following form:

\[\frac{2}{\sin x \cos x} \gt 0, \Rightarrow \frac{4}{2\sin x \cos x} \gt 0, \Rightarrow \frac{4}{\sin2x} \gt 0.\]

Hence it follows that

\[\sin2x \gt 0.\]

This basic trig inequality has the following solution:

\[2\pi n \lt 2x \lt \pi + 2\pi n,\]

or

\[\pi n \lt x \lt \frac{\pi}{2} + \pi n,\;n \in \mathbb{Z}.\]

The set of solutions is shown in Figure 1.

Solution of the inequality sin2x > 0
Figure 1.

Example 4.

Solve the trigonometric inequality

\[\cos2x \ge \sin x.\]

Solution.

Remember that

\[\cos2x = \cos^2x - \sin^2x = 1 - 2\sin^2x.\]

So the inequality takes the form

\[1 - 2\sin^2x - \sin x \ge 0.\]

Multiply both sides by minus one:

\[2\sin^2x + \sin x - 1 \le 0.\]

Denoting \(\sin x = t\) we solve the quadratic inequality:

\[2t^2 + t - 1 \le 0,\]
\[D = 1^2 -4\cdot 2 \cdot \left({-1}\right) = 9, \Rightarrow t_{1,2} = \frac{-1 \pm \sqrt{9}}{2\cdot 2} = \frac{-1 \pm 3}{4} = -1,\frac{1}{2}.\]

Taking into account the coefficient in front of the leading term, the solution of the inequality has the form

\[t \in \left[{-1, \frac{1}{2}}\right].\]

So

\[\sin x \in \left[{-1, \frac{1}{2}}\right].\]

Depict this set of solutions on the unit circle:

Solution of the inequality -1 <= sinx <= 1/2
Figure 2.

Thus the solution of the inequality is contained in the sector

\[-\frac{7\pi}{6} \le x \le \frac{\pi}{6},\]

or in general

\[x \in \left[{-\frac{7\pi}{6} + 2\pi n, \frac{\pi}{6}} + 2\pi n\right],\;n \in \mathbb{Z}.\]

Example 5.

Solve the inequality

\[\sqrt{\cos 2x} \lt \frac{\sqrt{2}}{2}.\]

Solution.

This inequality is equivalent to the following system of inequalities:

\[\sqrt{\cos 2x} \lt \frac{\sqrt{2}}{2}, \Rightarrow \left\{ \begin{array}{l} \cos 2x \ge 0\\ \left({\sqrt{\cos 2x}}\right)^2 \lt \left({\frac{\sqrt{2}}{2}}\right)^2 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} \cos 2x \ge 0\\ \cos 2x \lt \frac{1}{2} \end{array} \right..\]

We begin with the first inequality:

\[\cos 2x \ge 0, \Rightarrow -\frac{\pi}{2} + 2\pi n \le 2x \le \frac{\pi}{2} + 2\pi n, \Rightarrow -\frac{\pi}{4} + \pi n \le x \le \frac{\pi}{4} + \pi n,\;n \in \mathbb{Z}.\]

On the unit circle, the solution to this inequality is represented by two sectors.

Solution of the inequality cos2x >= 0
Figure 3.

Consider the second inequality:

\[\cos 2x \lt \frac{1}{2},\]
\[\Rightarrow \arccos{\frac{1}{2}} + 2\pi k \lt 2x \lt 2\pi - \arccos{\frac{1}{2}} + 2\pi k,\]
\[\Rightarrow \frac{\pi}{3} + 2\pi k \lt 2x \lt \frac{5\pi}{3} + 2\pi k,\]
\[\Rightarrow \frac{\pi}{6} + \pi k \lt x \lt \frac{5\pi}{6} + \pi k,\]

where \(k \in \mathbb{Z}.\)

The solutions to this inequality look like this

Solution of the inequality cos2x < 1/2
Figure 4.

Let's now superimpose both pictures on one unit circle and determine the set of angles that satisfy both inequalities:

Solution of the inequality \sqrt(cos2x) < \sqrt(2)/2
Figure 5.

We see that the solution in the first and fourth quadrants is described by the set

\[x \in \left[{-\frac{\pi}{4}, -\frac{\pi}{6}}\right) \cup \left({\frac{\pi}{6}, \frac{\pi}{4}}\right].\]

Add periodic terms and write down the final answer:

\[x \in \left[{-\frac{\pi}{4} + \pi m, -\frac{\pi}{6}} + \pi m\right) \cup \left({\frac{\pi}{6} + \pi\ell, \frac{\pi}{4} + \pi\ell}\right],\]

where \(m, \ell \in \mathbb{Z}.\)

Example 6.

Solve the trigonometric inequality

\[\frac{2\sin x - 1}{2\cos x - \sqrt{3}} \ge 0.\]

Solution.

Find the zeros of the numerator in the interval \(\left[{0, 2\pi}\right]:\)

\[2\sin x - 1 = 0, \Rightarrow \sin x = \frac{1}{2}, \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}.\]

Note that \(\sin x \ge \frac{1}{2}\) when \(x \in \left[{\frac{\pi}{6}, \frac{5\pi}{6}}\right].\)

We also examine the zeros of the denominator. Naturally, they do not belong to the domain of the inequality.

\[2\cos x - \sqrt{3} = 0, \Rightarrow \cos x = \frac{\sqrt{3}}{2}, \Rightarrow x = \pm\frac{\pi}{6}.\]

Draw all found roots on the unit circle:

Method of intervals on the unit trig circle
Figure 6.

At \(x = \frac{\pi}{6},\) both the numerator and denominator are zero, i.e. the point \(x = \frac{\pi}{6}\) has multiplicity 2. The inequality does not change its sign when passing through this point.

Determine the sign of the expression for \(x = 0:\)

\[\frac{2\sin 0 - 1}{2\cos 0 - \sqrt{3}} = \frac{- 1}{2 - \sqrt{3}} \lt 0.\]

Using the method of intervals, we put the signs on the unit circle, taking into account the multiplicity of the roots. As can be seen from the figure, the inequality has a solution in the range of angles \(\left[{\frac{5\pi}{6}, \frac{11\pi}{6}}\right).\) The final answer looks like

\[x \in \left[{\frac{5\pi}{6} + 2\pi n, \frac{11\pi}{6} + 2\pi n}\right),\;n \in \mathbb{Z}.\]