Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{4}+ k\pi\\x =\dfrac{\pi}{2} + k2\pi\\x = -\pi+ k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\cos2x + (1+2\cos x)(\sin x -\cos x) = 0$
$\Leftrightarrow \cos^2x -\sin^2x + (1+2\cos x)(\sin x -\cos x) = 0$
$\Leftrightarrow (\cos x -\sin x)(\cos x +\sin x) + (1+2\cos x)(\sin x -\cos x) = 0$
$\Leftrightarrow (\cos x -\sin x)(\cos x +\sin x - 1 - 2\cos x x)=0$
$\Leftrightarrow (\cos x -\sin x)(\sin x + \cos x + 1)=0$
$\Leftrightarrow \left[\begin{array}{l}\cos x -\sin x = 0\\\sin x +\cos x = -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sqrt2\cos\left(x +\dfrac{\pi}{4}\right) = 0\\\sqrt2\cos\left(x +\dfrac{\pi}{4}\right) =-1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\cos\left(x +\dfrac{\pi}{4}\right) = 0\\\cos\left(x +\dfrac{\pi}{4}\right) =-\dfrac{\sqrt2}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x +\dfrac{\pi}{4}= \dfrac{\pi}{2} + k\pi\\x +\dfrac{\pi}{4}=\dfrac{\sqrt3\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = -\dfrac{3\pi}{4} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{4}+ k\pi\\x =\dfrac{\pi}{2} + k2\pi\\x = -\pi+ k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
