Topic 3.10 Trigonometric Equations and Inequalities: solve trigonometric equations and inequalities, using inverse functions, symmetry and periodicity to find all solutions.

What this topic is asking

The College Board (Topic 3.10) wants you to solve trigonometric equations and inequalities, finding not just one solution but all of them. You use an inverse trig function to get a starting angle, the symmetry of the unit circle to find the second solution within a cycle, and periodicity to generate the complete set.

Solving the equation: one angle, then all

The inverse function (Topic 3.9) supplies the first angle; symmetry and periodicity supply the rest.

Why two solutions per cycle

For most values, a horizontal line crosses one cycle of sine or cosine twice. For sine, the two crossings are symmetric about $x = \frac{\pi}{2}$, giving an angle and its supplement $\pi - \theta$. For cosine, they are symmetric about the $x$-axis, giving $\theta$ and $-\theta$ (or $2\pi - \theta$). Tangent, with period $\pi$, has only one solution per cycle.

Trigonometric inequalities

To solve an inequality such as $\sin x > \frac{1}{2}$, first solve the equation $\sin x = \frac{1}{2}$ to find the boundary angles ($\frac{\pi}{6}$ and $\frac{5\pi}{6}$ on $[0, 2\pi)$). These split the interval into pieces; test one point in each piece to see where the inequality holds. Here $\sin x > \frac{1}{2}$ on $\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)$, the stretch between the boundary angles where the sine graph sits above the line $y = \frac{1}{2}$. This is the same sign-analysis method used for polynomial and rational inequalities in Unit 1.

A point worth stating once is that the most common error is stopping at the principal value. The inverse function gives only one angle; the exam almost always wants the second angle in the cycle and often the full general solution. Building the habit of asking "what is the other solution this cycle, and what does periodicity add?" after every inverse keeps you from losing easy marks.

Try this

Q1. Solve $\tan x = 1$ for all real $x$. [1 point]

• Cue. $\arctan(1) = \frac{\pi}{4}$; tangent has period $\pi$, so $x = \frac{\pi}{4} + \pi k$.

Q2. How many solutions does $\cos x = -1$ have on $[0, 2\pi)$? [1 point]

• Cue. One: $x = \pi$. The extreme values $\pm 1$ are hit only once per cycle.