Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
{\cos ^2}x + 4\sin x - 4 = 0\\
\Leftrightarrow \left( {1 - {{\sin }^2}x} \right) + 4\sin x - 4 = 0\\
\Leftrightarrow - {\sin ^2}x + 4\sin x - 3 = 0\\
\Leftrightarrow {\sin ^2}x - 4\sin x + 3 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = 3\,\,\,\,\,\left( {L, - 1 \le \sin x \le 1} \right)
\end{array} \right.\\
\Leftrightarrow \sin x = 1\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \,\,\,\,\,\,\left( {k \in Z} \right)\\
*)\\
y = 3\sin 2x - 5\\
- 1 \le \sin 2x \le 1\\
\Leftrightarrow - 3 \le 3\sin 2x \le 3\\
\Leftrightarrow - 8 \le 3\sin 2x - 5 \le - 2\\
\Leftrightarrow - 8 \le y \le - 2\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 8 \Leftrightarrow \sin 2x = - 1 \Leftrightarrow 2x = - \dfrac{\pi }{2} + k2\pi \Leftrightarrow x = - \dfrac{\pi }{4} + k\pi \\
{y_{\max }} = - 2 \Leftrightarrow \sin 2x = 1 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \Leftrightarrow x = \dfrac{\pi }{4} + k\pi
\end{array} \right.
\end{array}\)
