a)\(-1\le sinx\le1\)

\(\Leftrightarrow1\ge-sinx\ge-1\)

\(\Leftrightarrow4\ge3-sinx\ge2\) \(\Leftrightarrow16\ge\left(3-sinx\right)^2\ge4\)\(\Leftrightarrow17\ge\left(3-sinx\right)^2+1\ge5\)

\(\Leftrightarrow17\ge y\ge5\)

\(y_{min}=5\Leftrightarrow sinx=1\)\(\Leftrightarrow\)\(x=\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)

\(y_{max}=17\Leftrightarrow\)\(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)

b)\(y=\left(sin^2x+cos^2x\right)^2-2.sinx^2cos^2x\)\(=1-\dfrac{1}{2}.sin^22x\)

Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{1}{2}.sin^22x\ge-\dfrac{1}{2}\)

\(\Leftrightarrow1\ge1-\dfrac{1}{2}.sin^22x\ge\dfrac{1}{2}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{2}\)

\(y_{min}=\dfrac{1}{2}\Leftrightarrow sin^22x=1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}sin2x=-1\\sin2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)

c)\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-3sin^2x.cos^2x=1-\dfrac{3}{4}.sin^22x\)

Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{3}{4}.sin^22x\ge-\dfrac{3}{4}\)

\(\Leftrightarrow1\ge1-\dfrac{3}{4}.sin^22x\ge\dfrac{1}{4}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{4}\)

​\(y_{min}=\dfrac{1}{4}\Leftrightarrow sin^22x=1\)​\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)​​

\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)

Vậy...

a, Đặt \(t=sinx\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=\left(3-t\right)^2+1=t^2-6t+10\)

\(\Rightarrow min=min\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(1\right)=5\)

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(-1\right)=17\)

b, \(y=sin^4x+cos^4x=1-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)
Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=1-\dfrac{1}{2}t^2\)

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

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)

c, \(y=sin^6x+cos^6x\)

\(=sin^4x+cos^4x-sin^2x.cos^2x\)

\(=1-3sin^2x.cos^2x\)

\(=1-\dfrac{3}{4}sin^22x\)

Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=1-\dfrac{3}{4}t^2\)

\(\Rightarrow min=min\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=\dfrac{1}{4}\)

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)

2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$

\(ĐK:sinx-cosx\ne-2\)

\(< =>2y-1=sinx\left(1-y\right)+cosx\left(y+3\right)\)

Theo Bunhiacopxki:

\(\left[sinx\left(1-y\right)+cosx\left(y+3\right)\right]^2\)\(\le\left(sin^2x+cos^2x\right)\left[\left(1-y\right)^2+\left(y+3\right)^2\right]\)

\(< =>\left(2y-1\right)^2\le2y^2+4y+10\)

\(< =>2y^2-8y-9\le0\)

=> Bấm máy tìm Max, Min của y

(Sry máy tính của t bị ngáo không bấm ra)

\(\Rightarrow y.sinx-y.cosx+2y=sinx+3cosx+1\)

\(\Rightarrow\left(y-1\right)sinx-\left(y+3\right)cosx=1-2y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất

\(\Rightarrow\left(y-1\right)^2+\left(y+3\right)^2\ge\left(1-2y\right)^2\)

\(\Leftrightarrow2y^2-8y-9\le0\)

\(\Rightarrow\dfrac{4-\sqrt{34}}{2}\le y\le\dfrac{4+\sqrt{34}}{2}\)

\(y_{max}=\dfrac{4+\sqrt{34}}{2}\) ; \(y_{min}=\dfrac{4-\sqrt{34}}{2}\)

\(y=\dfrac{sinx+3cosx+1}{sinx-cosx+2}\)

\(\Leftrightarrow y.sinx-y.cosx+2y=sinx+3cosx+1\)

\(\Leftrightarrow\left(y-1\right)sinx-\left(y+3\right).cosx=1-2y\)

Phương trình có nghiệm khi \(\left(y-1\right)^2+\left(y+3\right)^2\ge\left(1-2y\right)^2\)

\(\Leftrightarrow y^2-2y+1+y^2+6y+9\ge4y^2-4y+1\)

\(\Leftrightarrow2y^2-8y-9\le0\)

\(\Leftrightarrow\dfrac{4-\sqrt{34}}{2}\le y\le\dfrac{4+\sqrt{34}}{2}\)

1. Không dịch được đề

2.

\(-1\le cos2x\le1\Rightarrow1\le y\le3\)

3.

a. \(-2\le2sinx\le2\Rightarrow-1\le y\le3\)

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

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

b.

\(0\le cos^2x\le1\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(cos^2x=1\Rightarrow x=k\pi\)

\(y_{max}=2\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

4.

\(y=\left(tanx-1\right)^2+2\ge2\)

\(y_{min}=2\) khi \(tanx=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

1.

\(y=\sqrt[4]{sinx}-\sqrt{cosx}\le\sqrt[4]{sinx}\le1\)

\(y_{max}=1\) khi \(\left\{{}\begin{matrix}sinx=1\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)

\(y=\sqrt[4]{sinx}-\sqrt{cosx}\ge-\sqrt{cosx}\ge-1\)

\(y_{min}=-1\) khi \(x=k2\pi\)

2.

\(y_{max}\) ko tồn tại

\(y=\frac{1}{cos^4x}+\frac{\sqrt{2}^2}{1-cos^4x}\ge\frac{\left(1+\sqrt{2}\right)^2}{cos^4x+1-cos^4x}=3+2\sqrt{2}\)

\(y_{min}=3+2\sqrt{2}\) khi \(cos^4x=\sqrt{2}-1\)