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a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)
Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=3-t^2\)
\(\Rightarrow y_{min}=minf\left(t\right)=2\)
\(y_{max}=maxf\left(t\right)=3\)
a.
\(0\le sin^2x\le1\Rightarrow\frac{4}{3}\le y\le4\)
\(y_{max}=4\) khi \(sinx=0\)
\(y_{min}=\frac{4}{3}\) khi \(sin^2x=1\)
b.
Đặt \(4sinx-3cosx=5\left(\frac{4}{5}sinx-\frac{3}{5}cosx\right)=5sin\left(x-a\right)=t\)
\(\Rightarrow-5\le t\le5\)
\(\Rightarrow y=t^2-4t+1=\left(t-2\right)^2-3\ge-3\)
\(y_{min}=-3\) khi \(t=2\)
\(y=t^2-4t-45+46=\left(t-9\right)\left(t+5\right)+46\le46\)
\(y_{max}=46\) khi \(t=-5\)
\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)
\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)
Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(y=\sqrt{5}sin\left(2x-a\right)+7\)
\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5sin^2x+1}=a\\\sqrt{5cos^2x+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}1\le a;b\le\sqrt{6}\\a^2+b^2=5\left(sin^2x+cos^2x\right)+2=7\end{matrix}\right.\)
\(y=a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{14}\)
\(y_{max}=\sqrt{14}\) khi \(cos2x=0\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Do \(1\le a\le\sqrt{6}\Rightarrow\left(a-1\right)\left(a-\sqrt{6}\right)\le0\)
\(\Rightarrow a\ge\dfrac{a^2+\sqrt[]{6}}{\sqrt{6}+1}\)
Tương tự ta có \(b\ge\dfrac{b^2+\sqrt{6}}{\sqrt{6}+1}\)
\(\Rightarrow y=a+b\ge\dfrac{a^2+b^2+2\sqrt{6}}{\sqrt{6}+1}=\dfrac{7+2\sqrt{6}}{\sqrt{6}+1}=\sqrt{6}+1\)
\(y_{min}=\sqrt{6}+1\) khi \(sin2x=0\Rightarrow x=\dfrac{k\pi}{2}\)
21.
a) `2sin(x-30^@)-1=0`
`<=>sin(x-30^@)=1/2`
`<=> sin(x-30^@)=sin30^@`
`<=>[(x-30^@=30^@+k360^@),(x-30^@=180^@-30^@+k360^@):}`
`<=> [(x=60^@+k360^@),(x=180^@+k360^@):}`
b) `5sin^2x+3cosx+3=0`
`<=>5(1-cos^2x)+3cosx+3=0`
`<=>-5cos^2x+3cosx+8=0`
`<=>(cosx+1)(cosx=8/5)=0`
`<=>[(cosx=-1),(cosx=8/5\ (VN)):}`
`<=>x=180^@+k360^@`
22.
`-1<=sin2x<=1`
`<=>2<=3+sin2x<=4`
`=> y_(min)=2 ; y_(max)=4`
TXĐ: D=R
y=4sin2x-4sinx+1+2
=(2sinx-1)2+2
Ta có:
-1\(\le\)sinx\(\le\)1
<=>-2\(\le\)2sinx\(\le\)2
<=>-3\(\le\)2sinx-1\(\le\)1
<=>0\(\le\)(2sinx-1)2\(\le\)1
<=>2\(\le\)(2sinx-1)2+2\(\le\)3
<=>2\(\le\)y\(\le\)11
=>Maxy=3<=>sinx=1<=>x=\(\dfrac{\Pi}{2}\)+k2\(\Pi\)
Miny=2<=>sinx=1/2<=>\(\left[{}\begin{matrix}x=\dfrac{\Pi}{6}+k2\Pi\\x=\dfrac{5\Pi}{6}+k2\Pi\end{matrix}\right.\)
à sai 1 chỗ là 2\(\le\)y\(\le\)3 nhé sửa lại giùm