Giải thích các bước giải:

$\begin{array}{l}
C1:\\
y = \sqrt 3 \sin x + \cos x + 2\\
 \Leftrightarrow \frac{y}{2} = \frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x + 1\\
 \Leftrightarrow \frac{y}{2} = \sin \left( {x + \frac{\pi }{6}} \right) + 1
\end{array}$

Mà $ - 1 \le \sin \left( {x + \frac{\pi }{6}} \right) \le 1$

$\begin{array}{l}
 \Leftrightarrow 0 \le \sin \left( {x + \frac{\pi }{6}} \right) + 1 \le 2\\
 \Leftrightarrow 0 \le \frac{y}{2} \le 2\\
 \Leftrightarrow 0 \le y \le 4
\end{array}$

+) Nếu $y=4$ 

$ \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = 1 \Leftrightarrow x + \frac{\pi }{6} = \frac{\pi }{2} + k2\pi  \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)$

$\to Maxy = 4 \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)$

+) Nếu $y=0$ 

$ \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) =  - 1 \Leftrightarrow x + \frac{\pi }{6} =  - \frac{\pi }{2} + k2\pi  \Leftrightarrow x =  - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right)$

$ \Rightarrow Miny = 0 \Leftrightarrow x =  - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right)$

Vậy 

$Maxy = 4 \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)$

$Miny = 0 \Leftrightarrow x =  - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right)$

C2:

Đường thẳng $\left( d \right):2x + y + 3 = 0$

Lấy điểm $A\left( { - 1, - 1} \right) \in \left( d \right)$

${T_{\overrightarrow v  = \left( {5, - 1} \right)}}A = A' \Rightarrow A'\left( {4, - 2} \right)$

Và ${T_{\overrightarrow v  = \left( {5, - 1} \right)}}\left( d \right) = \left( d \right)' \Rightarrow \left\{ \begin{array}{l}
A'\left( {4, - 2} \right) \in \left( d \right)'\\
\left( d \right)'//\left( d \right)
\end{array} \right.$

$\begin{array}{l}
 \Rightarrow \left( d \right)':2\left( {x - 4} \right) + 1\left( {y + 2} \right) = 0\\
 \Rightarrow \left( d \right)':2x + y - 6 = 0
\end{array}$

Vậy $\left( d \right)':2x + y - 6 = 0$