Đáp án:
$ \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \pm \dfrac{\pi}{12} + k\dfrac{\pi}{2}\end{array}\right.\qquad(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin^22x + \sin^24x = \sin^26x\\ \Leftrightarrow \sin^22x + 4\sin^22x.\cos^22x - (3\sin2x-4\sin^32x)^2 = 0\\ \Leftrightarrow \sin^22x + 4\sin^22x(1 - \sin^22x) - (9\sin^22x - 24\sin^42x + 16\sin^62x) = 0\\ \Leftrightarrow 4\sin^62x - 5\sin^42x +1\sin^22x = 0\\ \Leftrightarrow \sin^22x(4\sin^42x - 5\sin^22x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin^22x = 0\\\sin^22x = 1\\\sin^22x = \dfrac{1}{4}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\cos4x = -1\\\cos4x = \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \pm \dfrac{\pi}{12} + k\dfrac{\pi}{2}\end{array}\right.\qquad(k \in \Bbb Z) \end{array}$
