Đáp án:
$x=k\pi||x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}$
Giải thích các bước giải:
$cos^4x-cos2x-2sin^6x=0$
$\rightarrow (1-sin^2x)^2-(1-2sin^2x)-2sin^6x=0$
$\rightarrow sin^4x-2sin^2x+1-1+2sin^2x-2sin^6x=0$
$\rightarrow sin^4x-2sin^6x=0$
$\rightarrow sin^4x(1-2sin^2x)=0$
$\rightarrow sinx=0\quad||\quad sinx=\dfrac{1}{\sqrt[]{2}}\quad||\quad sinx=\dfrac{-1}{\sqrt[]{2}}$
$\rightarrow x=k\pi||x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}$
