Trig Equations and Inequalities — AP® Precalculus Notes

Solving $\sin x = \frac{1}{2}$ looks like a one-step job: press $\sin^{-1}$ . But sine repeats, so the equation has infinitely many solutions, and the inverse key returns only one of them. This page gives a four-step routine that finds every solution in an interval, works for equations like $\sin(3x) = 1$ and $\sin^2 x = \frac{1}{4}$ , and handles inequalities.

The routine: isolate, principal value, symmetry, period

Isolate the trigonometric function: undo the outside operations until you have $\sin(\ldots) = k$ , $\cos(\ldots) = k$ or $\tan(\ldots) = k$ . If $|k| > 1$ for sine or cosine, stop: there is no solution.
Principal value: apply the inverse (or read the unit circle) to get one angle.
Symmetry: find the other angle in the same period with the same value (Figure 1).
- Sine: $\sin(\pi - \theta) = \sin\theta$ , a mirror image across the $y$ -axis.
- Cosine: $\cos(-\theta) = \cos\theta$ , a mirror image across the $x$ -axis, so $2\pi - \theta$ also works.
- Tangent: one angle per period; the next is $\theta + \pi$ .
Period: add or subtract the period ( $2\pi$ for sine and cosine, $\pi$ for tangent) to reach every solution in the interval. Drop anything outside it.

A unit circle with the horizontal line y = 1/2 crossing it at two points: one on the terminal ray of pi/6 in Quadrant I and one on the terminal ray of 5 pi/6 in Quadrant II. The two rays are mirror images across the y-axis. — **Figure 1.** The horizontal line $y = \frac{1}{2}$ meets the unit circle at two angles. LevlPrep original diagram.

Worked example 1. Solve $2\cos x - \sqrt{3} = 0$ on $[0, 2\pi)$ .

Isolate: $2\cos x = \sqrt{3}$ , so $\cos x = \frac{\sqrt{3}}{2}$ .

Principal value: $\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$ .

Symmetry: $-\frac{\pi}{6}$ has the same cosine; it is outside the interval, so add $2\pi$ : $\frac{11\pi}{6}$ .

Answer: $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$ . On all real numbers: $x = \pm\frac{\pi}{6} + 2k\pi$ .

When the angle is $bx$

For $\sin(bx) = k$ , the inside $bx$ runs through a stretched interval. If $x$ is in $[0, 2\pi)$ and $b = 3$ , then $3x$ is in $[0, 6\pi)$ : three periods, so up to six solutions. Solve for the inside first, then divide.

Worked example 2. Solve $\sin(2x) = -\frac{\sqrt{2}}{2}$ on $[0, \pi)$ .

Let $u = 2x$ . Since $0 \le x < \pi$ , we have $0 \le u < 2\pi$ .

$\sin u = -\frac{\sqrt{2}}{2}$ : the principal value is $-\frac{\pi}{4}$ , outside $[0, 2\pi)$ , so use $-\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}$ . The mirror angle is $\pi - \left(-\frac{\pi}{4}\right) = \frac{5\pi}{4}$ .

So $u = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$ , and $x = \frac{u}{2} = \frac{5\pi}{8}$ or $\frac{7\pi}{8}$ . Both are in $[0, \pi)$ .

Quadratic forms and factoring

Treat $\sin x$ like a variable. $2\sin^2 x - \sin x - 1 = 0$ factors as $(2\sin x + 1)(\sin x - 1)$ $= 0$ , so $\sin x = -\frac{1}{2}$ or $\sin x = 1$ , and each piece is solved with the routine. Two rules keep every solution:

A square root has two signs: $\sin^2 x = \frac{1}{4}$ means $\sin x = \pm\frac{1}{2}$ .
Never divide both sides by a trigonometric expression that can be $0$ . Factor it out instead: $\tan x\sin x = \sin x$ becomes $\sin x(\tan x - 1) = 0$ , which keeps $x = 0$ and $x = \pi$ .

Inequalities

Solve the matching equation. Its solutions split the interval into pieces; on each piece the inequality is either true everywhere or false everywhere. Test one value per piece.

Worked example 3. Solve $\sin x < -\frac{1}{2}$ on $[0, 2\pi)$ .

$\sin x = -\frac{1}{2}$ at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ . Test $\frac{3\pi}{2}$ : $\sin\frac{3\pi}{2} = -1 < -\frac{1}{2}$ , true. Test $0$ : $\sin 0 = 0$ , false. So the solution is $\left(\frac{7\pi}{6}, \frac{11\pi}{6}\right)$ .

With a calculator

When the value is not on the unit circle, the routine still works with decimals. In radian mode, $\cos(2t) = \frac{1}{3}$ gives $2t$ $= \cos^{-1}\left(\frac{1}{3}\right)$ $\approx 1.230959$ ; store it. The other values are $2t = -1.230959 + 2k\pi$ and $2t = 1.230959 + 2k\pi$ . Divide by $2$ and keep the ones in the interval. To check, graph both sides and find every intersection, widening the window to the full interval so none is hidden.

Common slips

Stopping at the principal value and missing the second solution.
Listing an angle outside the interval, such as a negative principal value.
For $\sin(bx)$ , forgetting to stretch the interval, or forgetting to divide by $b$ at the end.