Đáp án:
Giải thích các bước giải:
$\quad \sin^2x + \sin^22x + \sin^23x + \sin^24x = 2$
$\Leftrightarrow \dfrac{1 -\cos2x}{2} +\dfrac{1-\cos4x}{2} + \dfrac{1-\cos6x}{2} + \dfrac{1-\cos8x}{2}= 2$
$\Leftrightarrow \cos2x +\cos4x + \cos6x +\cos8x = 0$
$\Leftrightarrow 2\cos3x.\cos x + 2\cos7x.\cos x = 0$
$\Leftrightarrow \cos x(\cos3x +\cos7x)= 0$
$\Leftrightarrow \cos x\cos2x\cos5x = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = 0\\\cos2x = 0\\\cos5x = 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \dfrac{\pi}{10} + k\dfrac{\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
