Đáp án:
$\left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x =k\dfrac{\pi}{9}\\x = -\dfrac{\pi}{2}+ k\pi\end{array}\right.\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin^23x + \sin^24x = \sin^25x+\sin^26x\\ \Leftrightarrow \dfrac{1 - \cos6x}{2}+\dfrac{1 - \cos8x}{2}=\dfrac{1 - \cos10x}{2}+\dfrac{1 - \cos12x}{2}\\ \Leftrightarrow (\cos6x - \cos10x) + (\cos8x - \cos12x) = 0\\ \Leftrightarrow -2\sin8x.\sin(-2x) - 2\sin10x.\sin(-2x) = 0\\ \Leftrightarrow \sin2x(\sin8x +\sin10x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\sin8x = \sin(-10x)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = k\pi\\8x = -10x + k2\pi\\8x = \pi + 10x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x =k\dfrac{\pi}{9}\\x = -\dfrac{\pi}{2}+ k\pi\end{array}\right.\,\,\,\,(k \in \Bbb Z)\\ \end{array}$
