Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{10} + k\dfrac{\pi}{5}\\x = \dfrac{\pi}{4} + \dfrac{\pi}{2}\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin^2x + \sin^22x +\sin^23x +\sin^24x = 2$
$\to \dfrac{1 -\cos2x}{2} + \dfrac{1 -\cos4x}{2}+ \dfrac{1 -\cos6x}{2}+ \dfrac{1 -\cos8x}{2} = 2$
$\to (\cos2x +\cos8x) +(\cos4x +\cos6x) = 0$
$\to 2\cos5x\cos3x +2\cos5x\cos x = 0$
$\to \cos5x(\cos3x +\cos x) = 0$
$\to \cos5x\cos2x\cos x = 0$
$\to \left[\begin{array}{l}\cos5x = 0\\\cos2x = 0\\\cos x = 0\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{10} + k\dfrac{\pi}{5}\\x = \dfrac{\pi}{4} + \dfrac{\pi}{2}\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
