Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = \dfrac{2\pi}{3} + k2\pi\\x = -\dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$1+\sin x+\cos x+\cos2x+\sin2x=0$
$\Leftrightarrow (\sin x + \cos x) + (1 + \sin2x) + \cos2x = 0$
$\Leftrightarrow (\sin x + \cos x) + (\sin x + \cos x)^2 + (\cos x + \sin x)(\cos x - \sin x) = 0$
$\Leftrightarrow (\sin x + \cos x)(1 + \sin x + \cos x + \cos x - \sin x) = 0$
$\Leftrightarrow (\sin x + \cos x)(2\cos x + 1) = 0$
$\Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right).(2\cos x + 1) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin\left(x + \dfrac{\pi}{4}\right) = 0\\\cos x = -\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = k\pi\\x = \dfrac{2\pi}{3} + k2\pi\\x = -\dfrac{2\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = \dfrac{2\pi}{3} + k2\pi\\x = -\dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
