Trigonometric Equations practice questions — AP Calculus BC

$\sin x = \cos x \iff \tan x = 1 \iff x = \pi/4 + k\pi$.

$2\cos^2 x + \cos x - 1 = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$. Roots: $\cos x = \tfrac{1}{2}$ ($x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$) and $\cos x = -1$ ($x = \pi$). Three solutions.

$2x = k\pi \Rightarrow x = k\pi/2$. On $[0, 2\pi)$: $0, \pi/2, \pi, 3\pi/2$. Four solutions.

$\cos^2 x = \tfrac{1}{2} \Rightarrow \cos x = \pm \tfrac{\sqrt{2}}{2}$. Solutions: $\tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$.

Factor as $(2\sin x + 1)(\sin x - 1) = 0$. $\sin x = 1$ gives $x = \pi/2$; $\sin x = -\tfrac{1}{2}$ gives $x = 7\pi/6,\, 11\pi/6$. Total: 3 solutions.

$\sin x (2\cos x - 1) = 0$. From $\sin x = 0$: $x = 0, \pi$. From $\cos x = \tfrac{1}{2}$: $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$. Total: $4$.

$\arctan(1) = \pi/4$. General solution: $x = \pi/4 + k\pi$ for any integer $k$.

$2u^2 - u - 1 = (2u + 1)(u - 1) = 0 \Rightarrow u = -1/2$ or $u = 1$. $\sin x = -1/2$: 2 solutions ($7\pi/6, 11\pi/6$). $\sin x = 1$: 1 solution ($\pi/2$). Total: $3$.

Squaring gives $\sin 2x = 0$, so $x \in \{0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}\}$. Substituting back into the original eliminates $\pi$ and $\tfrac{3\pi}{2}$. Solutions: $\{0, \tfrac{\pi}{2}\}$.

$\tan x = 1$ at $x = \tfrac{\pi}{4}$. Adding the period $\pi$ gives $x = \tfrac{5\pi}{4}$. Both lie in $[0, 2\pi)$.