Find the domain of the function.
$f(x)=({2+\cos 2x-\sin2x*\tan x})$^1/2
$2+\cos2x-\sin2x*\tan x≥0$
Simplifying
$-4\sin^2x+3≥0$
$\sin^2x≤3/4$
$-3/4≤\sin x≤3/4$
$\arcsin(-3/4)≤x≤\arcsin(3/4)$
My answer is - x ∈($\pi - \arcsin(3/4);\pi+\arcsin(3/4)∪(\arcsin(-3/4);\arcsin(3/4).$
Can someone check it, please?(i made an error during the last part, can someone check it with this error? xdd)
The last part is wrong, or at least incomplete. The condition you arrived at was equivalent to $|\sin x | \leq \frac{\sqrt{3}}{2}$. Take a look at $|\sin x|$... In $[0, 2\pi]$ this condition is satisfied for $x \in [0, \frac{\pi}{3}] \cup [\frac{2\pi}{3}, \frac{4 \pi}{3}] \cup [\frac{5\pi}{3}, 2 \pi] $. taking into account the period of $\sin$ the condition is satisfied in an infinite union of closed intervals given by $$ \bigcup_{k\in \mathbb{Z}} \left([2k\pi, \frac{\pi}{3}+2k\pi] \cup [\frac{2\pi}{3}+2k\pi, \frac{4 \pi}{3}+2k\pi] \cup [\frac{5\pi}{3}+2k\pi, 2 \pi+2k\pi]\right) $$