`~rai~`
\(f)\sin x+\cos x=-1\\\Leftrightarrow \sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x\right)=-1\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi.\end{array}\right.\\\text{Vậy S=}\left\{-\dfrac{\pi}{2}+k2\pi;\pi+k2\pi\Big|k\in\mathbb{Z}\right\}.\\g)\sin2x+\cos2x=\sqrt{2}\cos3x\\\Leftrightarrow \sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin2x+\dfrac{\sqrt{2}}{2}\cos2x\right)=\sqrt{2}\cos3x\\\Leftrightarrow \sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{2}\cos3x\\\Leftrightarrow \sin\left(2x+\dfrac{\pi}{4}\right)=\cos3x\\\Leftrightarrow \sin\left(2x+\dfrac{\pi}{4}\right)=\sin\left(\dfrac{\pi}{2}-3x\right)\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{4}=\dfrac{\pi}{2}-3x+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{2}+3x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}5x=\dfrac{\pi}{4}+k2\pi\\-x=\dfrac{\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{20}+k\dfrac{2\pi}{5}\\x=-\dfrac{\pi}{4}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{20}+k\dfrac{2\pi}{5};-\dfrac{\pi}{4}+k2\pi\Big|k\in\mathbb{Z}\right\}.\)
