$\begin{array}{l}
a)y = \sin x + \cos x\\
 \Leftrightarrow y = \sqrt 2 \left( {\dfrac{{\sqrt 2 }}{2}\sin x + \dfrac{{\sqrt 2 }}{2}\cos x} \right) = \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)\\
 - 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\
 \Leftrightarrow  - \sqrt 2  \le y \le \sqrt 2 \\
 \Rightarrow \left\{ \begin{array}{l}
\max y = \sqrt 2  \Rightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k2\pi \\
\min y =  - \sqrt 2  \Rightarrow x + \dfrac{\pi }{4} =  - \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \dfrac{{ - 3\pi }}{4} + k2\pi 
\end{array} \right.\\
b)y = \sqrt 3 \sin 2x - \cos 2x\\
y = 2\left( {\dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x} \right) = 2\sin \left( {2x - \dfrac{\pi }{6}} \right)\\
 - 1 \le \sin \left( {2x - \dfrac{\pi }{6}} \right) \le 1 \Rightarrow  - 2 \le y \le 2\\
 \Rightarrow \left\{ \begin{array}{l}
\max y = 2 \Rightarrow 2x - \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \dfrac{\pi }{3} + k\pi \\
\min y =  - 2 \Leftrightarrow 2x - \dfrac{\pi }{6} =  - \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x =  - \dfrac{\pi }{6} + k\pi 
\end{array} \right.
\end{array}$

`~rai~`

\(a)y=\sin x+\cos x\\\quad=\sqrt{2}.\left(\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x\right)\\\quad=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right).\\\text{Ta có:}-1\le\sin\left(x+\dfrac{\pi}{4}\right)\le 1\\\Leftrightarrow -\sqrt{2}\le\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\le\sqrt{2}\\\Leftrightarrow -\sqrt{2}\le y\le\sqrt{2}.\\+)Min=-\sqrt{2}\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=-1\\\Leftrightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=-\dfrac{3\pi}{4}+k2\pi.(k\in\mathbb{Z})\\+)Max_y=\sqrt{2}\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=1\\\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\\\text{Vậy Min}_y=-\sqrt{2}\text{ khi }x=-\dfrac{3\pi}{4}+k2\pi;\\\text{Max}_y=\sqrt{2}\text{ khi }x=\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\\b)y=\sqrt{3}\sin2x-\cos2x\\\quad=\dfrac{1}{2}.\left(\dfrac{\sqrt{3}}{2}\sin2x-\dfrac{1}{2}\cos2x\right)\\\quad=\dfrac{1}{2}\sin\left(2x-\dfrac{\pi}{6}\right).\\\text{Ta có:}-1\le\sin\left(2x-\dfrac{\pi}{6}\right)\le 1\\\Leftrightarrow -\dfrac{1}{2}\le\dfrac{1}{2}\sin\left(2x-\dfrac{\pi}{6}\right)\le\dfrac{1}{2}\\\Leftrightarrow -\dfrac{1}{2}\le y\le\dfrac{1}{2}.\\+)Min_y=-\dfrac{1}{2}\Leftrightarrow \sin\left(2x-\dfrac{\pi}{6}\right)=-1\\\Leftrightarrow 2x-\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow 2x=-\dfrac{\pi}{3}+k2\pi\\\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi.(k\in\mathbb{Z})\\+)Max_y=\dfrac{1}{2}\Leftrightarrow \sin\left(2x-\dfrac{\pi}{6}\right)=1\\\Leftrightarrow 2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow 2x=\dfrac{2\pi}{3}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{3}+k\pi.(k\in\mathbb{Z})\\\text{Vậy Min}_y=-\dfrac{1}{2}\text{ khi }x=-\dfrac{\pi}{6}+k\pi;\\\text{Max}_y=\dfrac{1}{2}\text{ khi }x=\dfrac{\pi}{3}+k\pi.(k\in\mathbb{Z})\)