Đáp án:
$\left[\begin{array}{l}x =k\dfrac{\pi}{2}\\x = \dfrac{\pi}{6}+ k\dfrac{\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\sin^2x + \sin^22x = \sin^23x$
$\Leftrightarrow 1 - \cos2x + 1 - \cos4x = 1 - \cos6x$
$\Leftrightarrow \cos6x - \cos4x + 1 - \cos2x = 0$
$\Leftrightarrow -2\sin5x\sin x + 2\sin^2x = 0$
$\Leftrightarrow \sin x(\sin x - \sin5x) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x = \sin5x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = 5x + k2\pi\\x = \pi - 5x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x =k\dfrac{\pi}{2}\\x = \dfrac{\pi}{6}+ k\dfrac{\pi}{3}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =k\dfrac{\pi}{2}\\x = \dfrac{\pi}{6}+ k\dfrac{\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$
