Đáp án + Giải thích các bước giải:
`sin4x=cos2x`
`<=>cos(4x-pi/2)=cos2x`
`<=>`\(\left[ \begin{array}{l}4x-\dfrac{\pi}{2}=2x+k2\pi\\4x-\dfrac{\pi}{2}=-2x+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k2\pi\\6x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{array} \right.\)`(kinZZ)`
