Trigonometric Equations

$\cos x=0⟺x=\pi \mathrm{/}2+k\pi$. On $[0,2\pi )$: $\pi \mathrm{/}2$ and $3\pi \mathrm{/}2$.

$\mathrm{\mid }\sin x\mathrm{\mid }\le 1$ for any real $x$, so $\sin x=2$ has no solutions.

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

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

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

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

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

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

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

$\sin x=\cos x⟺\tan x=1⟺x=\pi \mathrm{/}4+k\pi$.