Đáp án:
Giải thích các bước giải:
2cos4x-$\sqrt[]{3}$ = 0
⇔ 2cos4x = $\sqrt[]{3}$
⇔ cos4x = $\frac{√3}{2}$
⇔ cos4x = cos $\frac{ \pi }{6}$
⇔ \(\left[ \begin{array}{l}4x=\frac{\pi}{6}+k2\pi\\4x=-\frac{\pi}{6}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{\pi}{24}+\frac{k\pi}{2}\\x=-\frac{\pi}{24}+\frac{k\pi}{2} \end{array} \right.\)
Vậy: S={$\frac{\pi}{24}$ +$\frac{k\pi}{2}$ ; -$\frac{\pi}{24}$ +$\frac{k\pi}{2}$; k∈ Z }
