Đáp án A

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

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

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

\(-1\le sin\left(x^2\right)\le1\Rightarrow\)\(0\le\sqrt{1-sin\left(x^2\right)}\le\sqrt{2}\Rightarrow-1\le y\le\sqrt{2}-1\)

\(y_{min}=-1\) khi \(sin\left(x^2\right)=1\Rightarrow x=\pm\sqrt{\dfrac{\pi}{2}+k2\pi}\) (\(k\in N\))

\(y_{max}=\sqrt{2}-1\) khi \(sin\left(x^2\right)=-1\Rightarrow x=\pm\sqrt{-\dfrac{\pi}{2}+k2\pi}\) (\(k\in Z^+\))

\(sin\left(\dfrac{\pi}{3}+x\right)\in\left[-1;1\right]\)

\(\Rightarrow y=\dfrac{3}{2}+sin\left(\dfrac{\pi}{3}+x\right)\in\left[\dfrac{1}{2};\dfrac{5}{2}\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=\dfrac{1}{2}\\y_{max}=\dfrac{5}{2}\end{matrix}\right.\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

Ta có: \(0\le\sin^22x\le1\)

=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)

=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)

=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)

=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)

Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

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

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)

\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)

Ta có: \(0\le\sin^2x\le1\)

=>\(0\le6\sin^2x\le6\)

=>\(0-1\le6\sin^2x-1\le6-1\)

=>-1<=f(x)<=5

f(x) min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

f(x) max=5 khi \(\sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

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

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

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

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