Đáp án: \(\left[ \begin{array}{l}x =k\pi\\x =-\frac{\pi}{4}+k\pi\end{array} \right.\) (k∈Z)
Giải thích các bước giải:
Ta có:
cos(2x + $\frac{\pi}{4}$) = $\frac{\sqrt[]{2}}{2}$
⇔ cos(2x + $\frac{\pi}{4}$) = cos$\frac{\pi}{4}$
⇔ \(\left[ \begin{array}{l}2x + \frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x + \frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}2x =k2\pi\\2x =-\frac{\pi}{2}+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}x =k\pi\\x =-\frac{\pi}{4}+k\pi\end{array} \right.\) (k∈Z)
