Đáp án:
Giải thích các bước giải:
$cos4x-2sin2x-1=0$
⇔$(1-2sin^22x)-2sin2x-1=0$
⇔$-2sin^{2}2x-2sin2x=0$
⇔$-2sin2x^{}(sin2x+1)=0$
⇔\(\left[ \begin{array}{l}-2sin2x=0\\sin2x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}sin2x=0\\sin2x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=k\pi\\sin2x=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{k\pi}{2}\\x=\frac{-\pi}{4}+k\pi\end{array} \right.\)
