$ t = \sin x + \cos x\left( { - \sqrt 2  \le t \le \sqrt 2 } \right)\\  \Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x \Rightarrow {t^2} - 1 = \sin 2x\\ \sin 2x + 3\cos x + 3\sin x + 3 = 0\\  \Leftrightarrow {t^2} - 1 + 3t + 3 = 0 \Leftrightarrow {t^2} + 3t + 2 = 0 \Leftrightarrow \left( {t + 1} \right)\left( {t + 2} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} t =  - 1\\ t =  - 2(L) \end{array} \right. \Rightarrow t =  - 1 \Rightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  - 1\\  \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) =  - \dfrac{{\sqrt 2 }}{2}\\  \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} =  - \dfrac{\pi }{4} + k2\pi \\ x + \dfrac{\pi }{4} = \pi  + \dfrac{\pi }{4} + k2\pi  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x =  - \dfrac{\pi }{2} + k2\pi \\ x = \pi  + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right)\\ d) - \sin 2x - 3\sin x + 3\cos x + 3 = 0\\  \Leftrightarrow 3\sin x - 3\cos x + \sin 2x - 3 = 0\\ t = \sin x - \cos x\left( { - \sqrt 2  \le t \le \sqrt 2 } \right)\\  \Rightarrow {t^2} = 1 - 2\sin x\cos x = 1 - \sin 2x \Rightarrow \sin 2x = 1 - {t^2}\\ PT \Leftrightarrow 3t + 1 - {t^2} - 3 = 0 \Leftrightarrow {t^2} - 3t + 2 = 0 \Leftrightarrow \left( {t - 1} \right)\left( {t - 2} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t = 2(L) \end{array} \right. \Rightarrow \sin x - \cos x = 1 \Rightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 1\\  \Leftrightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2} \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{\pi }{4} = \dfrac{\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 }{2} + k2\pi \\ x = \pi  + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right) $

`~rai~`

\(c)\sin2x+3\cos x+3\sin x+3=0\\\Leftrightarrow 3(\sin x+\cos x)+2\sin x\cos x+3=0\quad (1)\\\text{Đặt }t=\sin x+\cos x\quad(|t|\le\sqrt{2})\\\Leftrightarrow t^2=(\sin x+\cos x)^2\\\Leftrightarrow t^2=\sin^2x+2\sin x\cos x+\cos^2x\\\Leftrightarrow t^2=1+2\sin x\cos x\\\Leftrightarrow 2\sin x\cos x=t^2-1.\\\text{Thay vào (1) được:}\\3t+t^2-1+3=0\\\Leftrightarrow t^2+3t+2=0\quad(a-b+c=0)\\\Leftrightarrow \left[\begin{array}{I}t=-1\text{(thỏa mãn)}\\t=-2\text{(loại)}\end{array}\right.\\\Leftrightarrow \sin x+\cos x=-1\\\Leftrightarrow \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=-1\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{-\dfrac{\pi}{2}+k2\pi;\pi+k2\pi\Big|k\in\mathbb{Z}\right\}.\\b)-\sin2x-3\sin x+3\cos x+3=0\\\Leftrightarrow 3\sin x-3\cos x+\sin2x-3=0\\\Leftrightarrow 3(\sin x-\cos x)+2\sin x\cos x-3=0\quad(1)\\\text{Đặt t=}\sin x-\cos x\quad(|t|\le\sqrt{2})\\\Leftrightarrow t^2=(\sin x-\cos x)^2\\\Leftrightarrow t^2=\sin^2x-2\sin x\cos x+\cos^2 x\\\Leftrightarrow t^2=1-2\sin x\cos x\\\Leftrightarrow 2\sin x\cos x=1-t^2.\\\text{Thay vào (1) được:}\\3t+1-t^2-3=0\\\Leftrightarrow t^2-3t+2=0\quad(a-b+c=0)\\\Leftrightarrow \left[\begin{array}{I}t=1\text{(thỏa mãn)}\\t=2\text{(loại)}\end{array}\right.\\\Leftrightarrow \sin x-\cos x=1\\\Leftrightarrow \sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)=1\\\Leftrightarrow \sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \left[\begin{array}{I}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{2}+k2\pi;\pi+k2\pi\Big|k\in\mathbb{Z}\right\}\)