a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

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Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

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Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

c.

$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$

$=\sin ^4x+2\sin ^2x-1$

$=(\sin ^4x-1)+(2\sin ^2x-2)+2$

$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$

$=(\sin ^2x-1)(\sin ^2x+3)+2$

Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$

$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$

$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$

Vậy $y_{\max}=2$

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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$

Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$

$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$

a: -1<=cos2x<=1

=>3>=-3cos2x>=-3

=>7>=-3cos2x+4>=1

=>7>=y>=1

\(y_{min}=1\) khi \(cos2x=1\)

=>2x=k2pi

=>x=kpi

\(y_{max}=-1\) khi cos2x=-1

=>2x=pi+k2pi

=>x=pi/2+kpi

b: \(0< =sin^2x< =1\)

=>\(3< =sin^2x+3< =4\)

=>3<=y<=4

y min=3 khi sin^2x=0

=>sinx=0

=>x=kpi

y max=4 khi sin^2x=1

=>cos^2x=0

=>x=pi/2+kpi

c: \(y=sin2x+3\)

-1<=sin2x<=1

=>-1+3<=sin2x+3<=1+3

=>2<=y<=4

\(y_{min}=2\) khi sin 2x=-1

=>2x=-pi/2+k2pi

=>x=-pi/4+kpi

y max=4 khi sin2x=1

=>2x=pi/2+k2pi

=>x=pi/4+kpi

a.

\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)

\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))

\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)

b.

\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)

\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)

\(\Leftrightarrow11y^2+2y-9\le0\)

\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)

c.

Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)

\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)

\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)

\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)

Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:

\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)

\(\Leftrightarrow y^2+8y-36\le0\)

\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)

a: Ta có: \(3\left|2x+5\right|\ge0\forall x\)

\(\Leftrightarrow3\left|2x+5\right|-7\ge-7\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)

c: ta có: \(\left(2x-3\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x-3\right)^2-14\ge-14\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

Dễ thấy D > 0

D có GTLN \(\Leftrightarrow\)( 2x - 3 )² + 5 có GTNN \(\Leftrightarrow\)( 2x - 3 )² có GTNN \(\Leftrightarrow\)2x = 3 \(\Leftrightarrow\)x = 1,5

GTLN của D = \(\frac{4}{5}\)khi x = 1,5

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