Đáp án:
a) $\left[\begin{array}{l}x = -\dfrac{\pi}{12}+ k\dfrac{2\pi}{3}\\x =\dfrac{3\pi}{4} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{6}\\x =\dfrac{\pi}{3} + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $\sin2x +\cos2x = -\sqrt2\sin x$
$\to \dfrac{\sqrt2}{2}\sin2x +\dfrac{\sqrt2}{2}\cos2x = -\sin x$
$\to \sin\left(2x +\dfrac{\pi}{4}\right) = \sin(-x)$
$\to \left[\begin{array}{l}2x +\dfrac{\pi}{4} = - x + k2\pi\\2x +\dfrac{\pi}{4} = \pi + x + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = -\dfrac{\pi}{12}+ k\dfrac{2\pi}{3}\\x =\dfrac{3\pi}{4} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\cos4x +\cos6x +\cos8x = 0$
$\to \cos6x + 2\cos6x\cos2x = 0$
$\to \cos6x(2\cos2x +1)=0$
$\to \left[\begin{array}{l}\cos6x = 0\\\cos2x = -\dfrac12\end{array}\right.$
$\to \left[\begin{array}{l}6x = \dfrac{\pi}{2}+ k\pi\\2x =\dfrac{2\pi}{3} + k2\pi\\2x = -\dfrac{2\pi}{3} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{6}\\x =\dfrac{\pi}{3} + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
