\(y=4-\frac{5}{4}\left(2sin2x.cos2x\right)^2\)

\(y=4-\frac{5}{4}sin^24x\)

Do \(0\le sin^24x\le1\)

\(\Rightarrow\frac{11}{4}\le y\le4\)

\(y_{min}=\frac{11}{4}\) khi \(sin^24x=1\)

\(y_{max}=4\) khi \(sin^24x=0\)

`TXĐ: R`

Ta có: `-1 <= sin(x+ \pi/3) <= 1`

`<=>0 <= sin^4 (x+\pi/3) <= 1`

`<=>2 <= y <= 3`

    `=>y_[mi n]=2<=>sin(x +\pi/3)=0<=>x= -\pi/3+k\pi`   `(k in ZZ)`

        `y_[max]=3<=>sin(x +\pi/3)=1<=>x=\pi/6 +k2\pi`  `(k in ZZ)`

Đáp án đúng : D

\(y=sin^3x+2sin^2x+sinx-2\)

đặt \(t=sinx\) với \(t\in\left[-1;1\right]\)

 pt \(\Leftrightarrow\)\(y=t^3+2t^2+t-2\)

\(y'=3t^2+4t+1\)

\(y'=0\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=-\dfrac{1}{3}\end{matrix}\right.\)

vậy GTLN của y = 2 với t=1 \(\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

GTNN của y=-58/27  với \(t=-\dfrac{1}{3}\Leftrightarrow sinx=-\dfrac{1}{3}\Leftrightarrow x=sin^{-1}\left(-\dfrac{1}{3}\right)\)

1.

\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{1}{2}(2\cos x\sin x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)

Dễ thấy:

$\sin ^22x\geq 0\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$

Vậy $y_{\max}=\sqrt{5}$

$\sin ^22x\leq 1\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\geq \sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$

Vậy $y_{\min}=\frac{3\sqrt{2}}{2}$

2.

$y=1+\frac{1}{2}\sin 2x\cos 2x=1+\frac{1}{4}.2\sin 2x\cos 2x$

$=1+\frac{1}{4}\sin 4x$

Vì $-1\leq \sin 4x\leq 1$

$\Rightarrow \frac{5}{4}\leq 1+\frac{1}{4}\sin 4x\leq \frac{3}{4}$

$\Leftrightarrow \frac{5}{4}\leq y\leq \frac{3}{4}$
Vậy $y_{\max}=\frac{5}{4}; y_{\min}=\frac{3}{4}$

3.

$\sin x\geq -1\Rightarrow \sqrt{1+\sin x}\geq 0$

$\Rightarrow y\geq -3$

Vậy $y_{\min}=-3$

$\sin x\leq 1\Rightarrow \sqrt{1+\sin x}\leq \sqrt{2}$

$\Rightarrow y\leq \sqrt{2}-3$
Vậy $y_{\max}=\sqrt{2}-3$

`y=sin^2 5x+cos^2 5x+3sin 2x-2`       `TXĐ: R`

`y=1+3sin 2x-2`

`y=3sin 2x-1`

Ta có: `-1 <= sin 2x <= 1`

`<=>-3 <= 3sin 2x <= 3`

`<=>-4 <= y <= 2`

    `=> y_[mi n]=4<=>sin 2x =-1<=>x=-\pi/4 + k\pi`  `(k in ZZ)`

         `y_[max] = 2<=>sin 2x=1<=>x=\pi/4+k\pi`  `(k in ZZ)`

a) \(y=1-2sinx\)

Ta có: \(-1\le sinx\le1\Rightarrow-2\le2sinx\le2\)

                                   \(\Rightarrow2\ge-2sin2x\ge-2\)

                                   \(\Rightarrow3\ge1-2sinx\ge-1\)

      Vậy \(y_{max}=3,y_{min}=-1\)