Thực hiện yêu cầu như đề bài trong hình

$\begin{array}{l}a)\,\,\sin2x + 2\cos x - \sin x - 1 = 0\\ \Leftrightarrow 2\sin x\cos x + 2\cos x - (\sin x + 1) = 0\\ \Leftrightarrow 2\cos x(\sin x + 1) - (\sin x + 1) = 0\\ \Leftrightarrow (\sin x +1)(2\cos x - 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = -1\\\cos x = \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,2\sin^2x - \sin2x + \sin x + \cos x = 1\\ \Leftrightarrow \sin x + \cos x = 1 + \sin2x - 2.\dfrac{1 - \cos2x}{2}\\ \Leftrightarrow \sin x + \cos x = \sin2x + \cos2x\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = \sin\left(2x + \dfrac{\pi}{4}\right)\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = 2x + \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} - 2x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = \dfrac{\pi}{6} +k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\\ c)\,\,\cos3x -2\sin2x - \cos x - \sin x = 1\\ \Leftrightarrow (\cos3x - \cos x)- 2\sin2x - \sin x - 1 = 0\\ \Leftrightarrow -2\sin2x.\sin x - 2\sin2x - (\sin x + 1) = 0\\ \Leftrightarrow -2\sin2x(\sin x + 1) - (\sin x + 1) = 0\\ \Leftrightarrow (\sin x + 1)(2\sin2x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = -1\\\sin2x = -\dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{2} + k2\pi\\x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{7\pi}{12} + k\pi\quad (k \in \Bbb Z)\end{array}\right.\\ d)\,\,\sin^2x + \sin x.\cos x - 2\cos^2x = 0\\ \text{Nhận thấy $\cos x =0$ không là nghiệm của phương trình}\\ \text{Chia 2 vế của phương trình cho $\cos^2x$ ta được:}\\ \tan^2x + \tan x - 2 =0\\ \Leftrightarrow \left[\begin{array}{l}\tan x = 1\\\tan x = -2\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} +k\pi\\ x = \arctan(-2) +k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$