Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
y = 2 + {\cos ^2}x - {\sin ^2}x\\
 = 2 + {\cos ^2}x - \left( {1 - {{\cos }^2}x} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\
 = 2 + {\cos ^2}x - 1 + {\cos ^2}x\\
 = 1 + 2{\cos ^2}x\\
 - 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\
 \Rightarrow 0 \le 2{\cos ^2}x \le 2\\
 \Rightarrow 1 \le 1 + 2{\cos ^2}x \le 1 + 2\\
 \Leftrightarrow 1 \le y \le 3\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow {\cos ^2}x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
{y_{\max }} = 3 \Leftrightarrow {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 0 \Rightarrow x = k\pi 
\end{array} \right.\\
3,\\
cos2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
y = {\cos ^2}x - 2\sqrt 3 \cos x.\sin x + 1\\
 = \dfrac{{\cos 2x + 1}}{2} - \sqrt 3 .\left( {2\cos x.\sin x} \right) + 1\\
 = \dfrac{1}{2}\cos 2x - \sqrt 3 \sin 2x + \dfrac{3}{2}\\
 - \sqrt {{a^2} + {b^2}}  \le a.\sin x + b.\sin y \le \sqrt {{a^2} + {b^2}} \\
 \Rightarrow  - \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}}  + \dfrac{3}{2} \le y \le \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}}  + \dfrac{3}{2}\\
 \Leftrightarrow \dfrac{{3 - \sqrt {13} }}{2} \le y \le \dfrac{{3 + \sqrt {13} }}{2}
\end{array}\)

$1) \, y = 2 + \cos^2x - \sin^2x$

$\Leftrightarrow y = 2 + \cos2x$

Ta có:

$-1 \leq \cos2x \leq 1$

$\Leftrightarrow 1 \leq 2 + \cos2x \leq 3$

Hay $1 \leq y \leq 3$

Vậy $\min y = 1 \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$

$\max y = 3 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$

$2)\, y = \cos^2x - 2\sqrt3\cos x\sin x + 1$

$\Leftrightarrow y = \dfrac{1 + \cos2x}{2} - \sqrt3\sin2x + 1$

$\Leftrightarrow y = \dfrac{1}{2}\cos2x - \sqrt3\sin2x + \dfrac{3}{2}$

$\Leftrightarrow y - \dfrac{3}{2} = \dfrac{1}{2}\cos2x - \sqrt3\sin2x$

Áp dụng bất đẳng thức $Bunyakovsky$ ta được:

(Phương trình có nghiệm)

$\left(y - \dfrac{3}{2}\right)^2 = \left(\dfrac{1}{2}\cos2x - \sqrt3\sin2x\right)^2 \leq \left(\dfrac{1}{4} + 3\right)(\cos^22x + \sin^22x) = \dfrac{13}{4}$

$\Rightarrow -\dfrac{\sqrt{13}}{2} \leq y - \dfrac{3}{2} \leq \dfrac{\sqrt{13}}{2}$

$\Leftrightarrow \dfrac{3 - \sqrt{13}}{2} \leq y \leq \dfrac{3 + \sqrt{13}}{2}$

Vậy $\min y = \dfrac{3 - \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x - \sqrt3\sin2x = - \dfrac{\sqrt{13}}{2} \Leftrightarrow x = \dfrac{\pi}{2} - \dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi$

$\max y = \dfrac{3 + \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x - \sqrt3\sin2x = \dfrac{\sqrt{13}}{2} \Leftrightarrow x = -\dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi \quad (k \in \Bbb Z)$