Đáp án:
$x = \left\{\dfrac{\pi}{12};\dfrac{7\pi}{12}\right\}$
Giải thích các bước giải:
$2\cos^22x + \sqrt3\sin4x = 3$
$\Leftrightarrow 1 + \cos4x + \sqrt3\sin4x = 3$
$\Leftrightarrow \dfrac{1}{2}\cos4x + \dfrac{\sqrt3}{2}\sin4x = 1$
$\Leftrightarrow \cos\left(4x - \dfrac{\pi}{3}\right) = 1$
$\Leftrightarrow 4x - \dfrac{\pi}{3} = k2\pi$
$\Leftrightarrow x = \dfrac{\pi}{12} + k\dfrac{\pi}{2}\quad (k\in \Bbb Z)$
Ta có:
$0 < x < \pi$
$\Leftrightarrow 0 < \dfrac{\pi}{12} + k\dfrac{\pi}{2} < \pi$
$\Leftrightarrow -\dfrac{1}{6} < k < \dfrac{11}{6}$
Do $k \in \Bbb Z$
nên $k = \left\{0;1\right\}$
$\Rightarrow x = \left\{\dfrac{\pi}{12};\dfrac{7\pi}{12}\right\}$
