Đáp án: $x\in\{\dfrac{\pi}{4}+k\pi,\dfrac{\pi}{8}+\dfrac{k\pi}{2}\}$
Giải thích các bước giải:
$\cos^3x+\sin x-3\sin^2x\cos x=0$
$\rightarrow \cos^3x+\sin x(\sin^2x+\cos^2x)-3\sin^2x\cos x=0$
$\rightarrow \cos^3x+\cos^2x\sin x-3\sin^2x\cos x+\sin^3x=0$
$\rightarrow \cos^3x+\cos^2x\sin x-3\sin^2x\cos x+\sin^3x=0$
$\rightarrow (\cos x-\sin x)(\cos^2x+2\sin x\cos x-sin^2x)=0$
$\rightarrow (\cos x-\sin x)(\cos 2x+\sin 2x)=0$
$+)\cos x=\sin x$
$\rightarrow \tan x=1$
$\rightarrow x=\dfrac{\pi}{4}+k\pi$
$+)\cos 2x=\sin 2x$
$\rightarrow \tan 2x=1$
$\rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}$
