Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{12} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$4\sin^22x = 1$
$\Leftrightarrow 2(1 - \cos4x) = 1$
$\Leftrightarrow \cos4x = \dfrac{1}{2}$
$\Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{3} + k2\pi\\4x = -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{12} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
