a. Ta có: $\sin x\in [-1;1]$ nên $|\sin x|\in [0;1]$
$\Rightarrow 1\leq 3-2|\sin x|\leq 3$
Vậy $y_{\min}=1; y_{\max}=3$
b.
$y=\frac{1-\cos 2x}{2}-\frac{3}{2}\sin 2x+1$
$2y=3-\cos 2x-3\sin 2x$
$3-2y=\cos 2x+3\sin x$
Áp dụng định lý Bunhiacopxky:
$(3-2y)^2\leq (\cos ^22x+\sin ^22x)(1+3^2)=10$
$\Rightarrow -\sqrt{10}\leq 3-2y\leq \sqrt{10}$
$\Rightarrow \frac{3-\sqrt{10}}{2}\leq y\leq \frac{3+\sqrt{10}}{2}$
Vậy $y_{\max}=\frac{1+\sqrt{10}}{2}; y_{\min}=\frac{1-\sqrt{10}}{2}$
c.
\(y=\sqrt{5-\frac{1}{4}(2\sin x\cos x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)
Vì $\sin 2x\in [-1;1]$
$\Rightarrow \sin ^22x\in [0;1]$
$\Rightarrow \frac{3\sqrt{2}}{2}\leq \sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$
d.
$\cos (x+\frac{\pi}{3})\in [-1;1]$
$\Rightarrow 2(-1)+3\leq 2\cos (x+\frac{\pi}{3})+3\leq 2.1+3$
$\Rightarrow 1\leq y\leq 5$
$\Rightarrow y_{\min}=1; y_{\max}=5$
e.
\(y=2\sin ^2x-\cos 2x=1-\cos 2x-\cos 2x=1-2\cos 2x\)
Vì $\cos 2x\in [-1;1]$ nên:
$-1\leq 1-2\cos 2x\leq 3$
Vậy $y_{\min}=-1; y_{\max}=3$
f.
$y=(\sin ^2x+\cos ^2x)^2-2(\sin x\cos x)^2$
$=1-\frac{1}{2}\sin ^22x$
Vì $-1\leq \sin 2x\leq 1\Rightarrow 0\leq \sin ^22x\leq 1$
$\Rightarrow \frac{1}{2}\leq y\leq 1$
Vậy $y_{\min}=\frac{1}{2};y_{\max}=1$
g. Áp dụng BĐT Bunhiacopxky:
$y^2=(\sin x+\sqrt{3}\cos x)^2\leq (\sin ^2x+\cos ^2x)(1+3)=4$
$\Rightarrow -2\leq y\leq 2$
$\Rightarrow y_{\min}=-2; y_{\max}=2$
h. Với $x\in [\frac{-\pi}{3}; \frac{\pi}{6}]$
$\Rightarrow \frac{1}{4}\leq \cos ^2x\leq 1$
$\Rightarrow 1\leq \frac{1}{\cos ^2x}\leq 4$
Vậy $y_{\min}=1; y_{\max}=4$
