Đáp án:

$\begin{array}{l}
Miny = \dfrac{3}{4} \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
Maxy = \dfrac{5}{4} \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}$

Giải thích các bước giải:

Bài 1:

$\begin{array}{l}
1)\cos 5x = \sin 3x\\
 \Leftrightarrow \cos 5x = \cos \left( {\dfrac{\pi }{2} - 3x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{\pi }{2} - 3x + k2\pi \\
5x =  - \dfrac{\pi }{2} + 3x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{4}\\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
2){\sin ^2}2x + {\sin ^2}3x = 1\\
 \Leftrightarrow {\sin ^2}2x = 1 - {\sin ^2}3x\\
 \Leftrightarrow {\sin ^2}2x = {\cos ^2}3x\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \cos 3x\\
\sin 2x =  - \cos 3x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \sin \left( {\dfrac{\pi }{2} - 3x} \right)\\
\sin 2x = \cos \left( {3x + \pi } \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \sin \left( {\dfrac{\pi }{2} - 3x} \right)\\
\sin 2x = \sin \left( { - \dfrac{\pi }{2} - 3x} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} - 3x + k2\pi \\
2x = \dfrac{\pi }{2} + 3x + k2\pi \\
2x =  - \dfrac{\pi }{2} - 3x + k2\pi \\
2x = \dfrac{{3\pi }}{2} + 3x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + k\dfrac{{2\pi }}{5}\\
x =  - \dfrac{\pi }{2} + k2\pi \\
x =  - \dfrac{\pi }{{10}} + k\dfrac{{2\pi }}{5}\\
x =  - \dfrac{{3\pi }}{2} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Bài 2:

Ta có:

$y = 1 + \dfrac{1}{2}\sin x.\cos x = 1 + \dfrac{1}{4}\sin 2x$

Mà $ - 1 \le \sin 2x \le 1 \Rightarrow \dfrac{{ - 1}}{4} \le \dfrac{1}{4}\sin 2x \le \dfrac{1}{4}$

$\begin{array}{l}
 \Rightarrow \dfrac{3}{4} \le y \le \dfrac{5}{4}\\
Miny = \dfrac{3}{4} \Leftrightarrow \sin 2x =  - 1 \Leftrightarrow 2x =  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right) \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
Maxy = \dfrac{5}{4} \Leftrightarrow \sin 2x = 1 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right) \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}$

Vậy $\begin{array}{l}
Miny = \dfrac{3}{4} \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
Maxy = \dfrac{5}{4} \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}$