a: \(-1\le\sin x\le1\)

=>\(-1+1\le\sin x+1\le1+1\)

=>\(0\le\sin x+1\le2\)

=>\(0\le3\left(\sin x+1\right)\le6\)

=>\(0\le\sqrt{3\left(\sin x+1\right)}\le\sqrt6\)

=>\(0-5\le\sqrt{3\left(\sin x+1\right)}-5\le\sqrt6-5\)

=>-5<=y<=\(\sqrt6-5\)

Do đó: \(y_{\min}=-5\) khi sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

\(y_{\max}=\sqrt6-5\) khi sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

b: \(-1\le\sin\left(x+8\right)\le1\)

=>\(-6\le6\sin\left(x+8\right)\le6\)

=>\(-6-5\le6\sin\left(x+8\right)-5\le6-5\)

=>-11<=y<=1

Vậy: \(y_{\min}=-11\) khi sin (x+8)=-1

=>\(x+8=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+k2\pi-8\)

\(y_{\max}=1\) khi sin(x+8)=1

=>\(x+8=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{2}+k2\pi-8\)

Đặt \(sin^24x=t\left(t\in\left[0;1\right]\right)\)

\(y=1-8sin^22x.cos^22x+2sin^42x\)

\(=1-2sin^24x+2sin^42x\)

\(\Rightarrow y=f\left(t\right)=1-2t+2t^2\)

\(y_{min}=min\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=\dfrac{1}{2}\)

\(y_{max}=max\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=1\)

a: \(-1\le\sin x\le1\)

=>\(-1+1\le\sin x+1\le1+1\)

=>\(0\le\sin x+1\le2\)

=>\(0\le6\left(\sin x+1\right)\le2\cdot6=12\)

=>\(0\le\sqrt{6\left(\sin x+1\right)}\le\sqrt{12}=2\sqrt3\)

=>\(0-9\le\sqrt{6\left(\sin x+1\right)}-9\le=2\sqrt3-9\)

=>\(-9\le y\le2\sqrt3-9\)

Do đó, ta có:

\(y_{\min}=-9\) khi sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

\(y_{\max}=2\sqrt3-9\) khi sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

b: \(-1\le\sin\left(x+1\right)\le1\)

=>\(-4\le4\sin\left(x+1\right)\le4\)

=>\(-4-7\le4\sin\left(x+1\right)-7\le4-7\)

=>-11<=y<=-3

Vậy: \(y_{\min}=-11\) khi sin(x+1)=-1

=>\(x+1=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+k2\pi-1\)

\(y_{\max}\) =-3 khi sin(x+1)=1

=>\(x+1=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{2}+k2\pi-1\)

Lời giải:

$y=2\sin ^2x+\sqrt{3}\sin 2x=1-\cos 2x+\sqrt{3}\sin 2x$

$=1-(\cos 2x-\sqrt{3}\sin 2x)$

Áp dụng BĐT Bunhiacopxky:

$(\cos 2x-\sqrt{3}\sin 2x)^2\leq (\cos ^22x+\sin ^22x)(1+3)=4$

$\Rightarrow \cos 2x-\sqrt{3}\sin 2x\leq 2$

$\Rightarrow y=1-(\cos 2x-\sqrt{3}\sin 2x)\geq -1$

Vậy $y_{\min}=-1$. Giá trị này đạt tại $x=\frac{5\pi}{6}+2k\pi$ hoặc $x=\frac{-\pi}{6}+2k\pi$ với $k$ nguyên bất kỳ.

Đặt \(sinx=t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^2+2t\)

Xét hàm \(y=f\left(t\right)=t^2+2t\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-1;1\right]\)

\(f\left(-1\right)=-1\) ; \(f\left(1\right)=3\)

\(\Rightarrow y_{min}=-1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=3\) khi \(sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)