Trigonometric Equations

$\cos x = 0 \iff x = \pi/2 + k\pi$. On $[0, 2\pi)$: $\pi/2$ and $3\pi/2$.

$|\sin x| \le 1$ for any real $x$, so $\sin x = 2$ has no solutions.

$x = \pi/6$ and $x = \pi - \pi/6 = 5\pi/6$. Two solutions.

Reference angle $\tfrac{\pi}{4}$. In $[0, 2\pi)$ cosine is $\tfrac{\sqrt{2}}{2}$ at $x = \tfrac{\pi}{4}$ (Q1) and $x = \tfrac{7\pi}{4}$ (Q4).

$\sin x = \tfrac{1}{2}$ at $x = \tfrac{\pi}{6}$ (Q1) and $x = \tfrac{5\pi}{6}$ (Q2). Two solutions on $[0, 2\pi)$.

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

$\sin x = \tfrac{\sqrt{3}}{2}$ at the reference angle $\tfrac{\pi}{3}$. Sine is positive in Q1 and Q2: $x = \tfrac{\pi}{3}$ and $x = \tfrac{2\pi}{3}$.

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.

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

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