Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
10,\\
y = {\sin ^4}x - 2{\cos ^2}x + 1\\
 = {\sin ^4}x - 2.\left( {1 - {{\sin }^2}x} \right) + 1\\
 = {\sin ^4}x - 2 + 2{\sin ^2}x + 1\\
 = \left( {{{\sin }^4}x + 2{{\sin }^2}x + 1} \right) - 2\\
 = {\left( {{{\sin }^2}x + 1} \right)^2} - 2\\
 - 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^2}x \le 1\\
 \Leftrightarrow 1 \le {\sin ^2}x + 1 \le 2\\
 \Leftrightarrow 1 \le {\left( {{{\sin }^2}x + 1} \right)^2} \le 4\\
 \Rightarrow  - 1 \le y \le 2\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} =  - 1 \Leftrightarrow {\sin ^2}x = 0 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = 2 \Leftrightarrow {\sin ^2}x = 1 \Leftrightarrow \sin x =  \pm 1 \Leftrightarrow x = \frac{\pi }{2} + k\pi 
\end{array} \right.\\
11,\\
y = {\sin ^4}x + {\cos ^4}x + 4\\
 = {\sin ^4}x + {\left( {1 - {{\sin }^2}x} \right)^2} + 4\\
 = {\sin ^4}x + \left( {1 - 2{{\sin }^2}x + {{\sin }^4}x} \right) + 4\\
 = 2{\sin ^4}x - 2{\sin ^2}x + 5\\
 = 2.\left( {{{\sin }^4}x - {{\sin }^2}x + \frac{1}{4}} \right) + \frac{9}{2}\\
 = 2.{\left( {{{\sin }^2}x - \frac{1}{2}} \right)^2} + \frac{9}{2}\\
 - 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^2}x \le 1\\
 \Leftrightarrow  - \frac{1}{2} \le {\sin ^2}x - \frac{1}{2} \le \frac{1}{2}\\
 \Leftrightarrow 0 \le {\left( {{{\sin }^2}x - \frac{1}{2}} \right)^2} \le \frac{1}{4}\\
 \Leftrightarrow \frac{9}{2} \le y \le 5\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \frac{9}{2} \Leftrightarrow {\sin ^2}x = \frac{1}{2} \Leftrightarrow \sin x = \frac{{ \pm \sqrt 2 }}{2} \Leftrightarrow x = \frac{\pi }{2} + \frac{{k\pi }}{2}\\
{y_{\max }} = 5 \Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 1\\
{\sin ^2}x = 0
\end{array} \right. \Rightarrow x = \frac{{k\pi }}{2}
\end{array} \right.\\
12,\\
y = {\sin ^6}x + {\cos ^6}x\\
 = \left( {{{\sin }^2}x + {{\cos }^2}x} \right).\left( {{{\sin }^4}x - {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)\\
 = {\sin ^4}x - {\sin ^2}x.{\cos ^2}x + {\cos ^4}x\\
 = {\sin ^4}x - {\sin ^2}x.\left( {1 - {{\sin }^2}x} \right) + {\left( {1 - {{\sin }^2}x} \right)^2}\\
 = {\sin ^4}x - {\sin ^2}x + {\sin ^4}x + 1 - 2{\sin ^2}x + {\sin ^4}x\\
 = 3{\sin ^4}x - 3{\sin ^2}x + 1\\
 = 3.\left( {{{\sin }^4}x - {{\sin }^2}x + \frac{1}{4}} \right) + \frac{1}{4}\\
 = 3.{\left( {{{\sin }^2}x - \frac{1}{2}} \right)^2} + \frac{1}{4}\\
 - 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^2}x \le 1\\
 \Leftrightarrow  - \frac{1}{2} \le {\sin ^2}x - \frac{1}{2} \le \frac{1}{2}\\
 \Leftrightarrow 0 \le {\left( {{{\sin }^2}x - \frac{1}{2}} \right)^2} \le \frac{1}{4}\\
 \Leftrightarrow \frac{1}{4} \le y \le 1\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \frac{1}{4} \Leftrightarrow {\sin ^2}x = \frac{1}{2} \Leftrightarrow \sin x = \frac{{ \pm \sqrt 2 }}{2} \Leftrightarrow x = \frac{\pi }{2} + \frac{{k\pi }}{2}\\
{y_{\max }} = 1 \Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 1\\
{\sin ^2}x = 0
\end{array} \right. \Rightarrow x = \frac{{k\pi }}{2}
\end{array} \right.\\
13,\\
y = \sin 2x + \sqrt 3 \cos 2x + 5\\
 = 2.\left( {\frac{1}{2}\sin 2x + \frac{{\sqrt 3 }}{2}\cos 2x} \right) + 5\\
 = 2.\sin \left( {2x + \frac{\pi }{3}} \right) + 5\\
 - 1 \le \sin \left( {2x + \frac{\pi }{3}} \right) \le 1 \Rightarrow 3 \le y \le 7\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 3 \Leftrightarrow \sin \left( {2x + \frac{\pi }{3}} \right) =  - 1\\
{y_{\max }} = 7 \Leftrightarrow \sin \left( {2x + \frac{\pi }{3}} \right) = 1
\end{array} \right.
\end{array}\)