Đáp án:
a) $\left[\begin{array}{l}x= -\dfrac{\pi}{4} +k2\pi\\x =\dfrac{\pi}{20}+ k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x =\dfrac{\pi}{10} + k\dfrac{\pi}{5}\\x =\dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x =\dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $\sin2x +\cos2x -\sqrt2\cos3x = 0$
$\to \dfrac{\sqrt2}{2}\sin2x +\dfrac{\sqrt2}{2}\cos2x =\cos3x$
$\to \cos\left(2x -\dfrac{\pi}{4}\right) =\cos3x$
$\to \left[\begin{array}{l}2x -\dfrac{\pi}{4}=3x +k2\pi\\2x -\dfrac{\pi}{4}=-3x + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x= -\dfrac{\pi}{4} +k2\pi\\x =\dfrac{\pi}{20}+ k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
b) $\sin^2x +\sin^22x +\sin^23x +\sin^24x =2$
$\to \dfrac{1 -\cos2x}{2}+\dfrac{1 -\cos4x}{2}+\dfrac{1-\cos6x}{2} +\dfrac{1 -\cos8x}{2}=2$
$\to \cos2x +\cos8x +\cos4x +\cos6x = 0$
$\to 2\cos5x.\cos3x +2\cos5x.\cos x = 0$
$\to \cos5x(\cos3x +\cos x) = 0$
$\to \cos5x.\cos2x.\cos x = 0$
$\to \left[\begin{array}{l}\cos5x=0\\\cos2x = 0\\\cos x = 0\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{10} + k\dfrac{\pi}{5}\\x =\dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x =\dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
