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

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

Tham khảo:

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

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

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

\(\Leftrightarrow17\ge y\ge5\)

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

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

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

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

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

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

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

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

\(=5-2\cdot\left\lbrack\frac12\cdot2\cdot\sin x\cdot cosx\right\rbrack^2=5-2\cdot\left\lbrack\frac12\cdot\sin2x\right\rbrack^2\)

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

\(0\le\sin^22x\le1\)

=>\(0\ge-\frac12\sin^22x\ge-\frac12\)

=>\(0+5\ge-\frac12\sin^22x+5\ge-\frac12+5\)

=>\(5\ge-\frac12\sin^22x+5\ge\frac92\)

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

=>\(\sqrt{\frac92}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

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

=>\(\frac{2}{3\sqrt2}\ge\frac{1}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1}{\sqrt5}\)

=>\(\frac{2\cdot4}{3\sqrt2}\ge\frac{1\cdot4}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1\cdot4}{\sqrt5}\)

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

=>\(y_{\max}=\frac{4\sqrt2}{3}\) khi \(-\frac12\cdot\sin^22x+5=\frac92\)

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

=>\(\sin^22x=1\)

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

=>cos2x=0

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

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

\(y_{\min}=\frac{4}{\sqrt5}\) khi \(-\frac12\cdot\sin^22x+5=5\)

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

=>sin 2x=0

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

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

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

\(=3\left(1-cos^2x\right)+5\cdot cos^2x-4\left(2\cdot cos^2x-1\right)-2\)

\(=3-3\cdot cos^2x+5\cdot cos^2x-8\cdot cos^2x+4-2=-6\cdot cos^2x+5\)

Ta có: \(0<=cos^2x\le1\)

=>\(0\ge-6\cdot cos^2x\ge-6\)

=>\(0+5\ge-6\cdot cos^2x+5\ge-6+5\)

=>5>=y>=-1

Do đó: \(y_{\min}=-1\) khi \(-6\cdot cos^2x+5=-1\)

=>\(-6\cdot cos^2x=-6\)

=>\(cos^2x=1\)

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

=>sin x=0

=>\(x=k\pi\)

y max=5 khi \(-6\cdot cos^2x+5=5\)

=>\(-6\cdot cos^2x=0\)

=>cosx=0

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

Đặt \(t=\sin^2x\Rightarrow\begin{cases}\cos^2x=1-t\\t\in\left[0;1\right]\end{cases}\) \(\Leftrightarrow f\left(x\right)=5^t+5^{1-t}=g\left(t\right);t\in\left[0;1\right]\)

Ta có : \(g'\left(t\right)=5^t\ln5-5^{1-t}\ln5=\left(5^t-5^{1-t}\right)\ln5=0\)

           \(\Leftrightarrow5^t=5^{1-t}\)

           \(\Leftrightarrow t=1-t\)

           \(t=\frac{1}{2}\)

Mà \(\lim\limits_{x\rightarrow-\infty}g\left(t\right)=\lim\limits_{x\rightarrow-\infty}\left(5^t-5^{1-t}\right)=+\infty\)

       \(\lim\limits_{x\rightarrow+\infty}g\left(t\right)=\lim\limits_{x\rightarrow+\infty}\left(5^t-5^{1-t}\right)=+\infty\)

Ta có bảng biến thiên

t g'(t) g(t) - 8 1 2 + 8 0 - + + 8 + 8 2 căn 5

\(\Rightarrow\) Min \(f\left(x\right)=2\sqrt{5}\) khi  \(t=\frac{1}{2}\Leftrightarrow\sin^2x=\frac{1}{2}\Leftrightarrow\frac{1-\cos2x}{2}=\frac{1}{2}\)

                                             \(\Leftrightarrow\cos2x=0\)                  

                                              \(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)   \(\left(k\in Z\right)\)