![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=2-2.\left(2sinx.cosx\right)=2-2sin2x\)
Do \(-1\le sin2x\le1\Rightarrow0\le y\le4\)
\(y_{min}=0\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)
\(y_{max}=4\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(cos2x\ge0\Rightarrow0\le cos2x\le1\)
\(\Rightarrow3.0+1\le y\le3.\sqrt{1}+1\)
\(\Rightarrow1\le y\le4\)
\(y_{min}=1\) khi \(cos2x=0\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
\(y_{max}=4\) khi \(cos2x=1\Rightarrow x=k\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(sin\left(x-\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)
\(\Rightarrow y=2sin\left(x-\dfrac{\pi}{3}\right)+3\in\left[1;5\right]\)
\(\Rightarrow\left\{{}\begin{matrix}y_{min}=1\\y_{max}=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(0\le cos^2x\le1\Rightarrow\dfrac{1}{2}\le y\le2\)
\(y_{min}=\dfrac{1}{2}\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)
\(y_{max}=2\) khi \(cos^2x=1\Rightarrow sinx=0\Rightarrow x=k\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)
Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)
\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)
\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)
\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)
\(y_{min}=0\) khi \(sinx=-1\)
Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)
Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)
Vậy \(y_{max}=\dfrac{97}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(sinx-3\right)\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=3>1\left(loại\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(y=2\left(1-cos2x\right)-cos2x=2-3cos2x\)
Do \(-1\le cos2x\le1\Rightarrow-1\le y\le5\)
\(y_{min}=-1\) khi \(cos2x=1\Leftrightarrow x=k\pi\)
\(y_{max}=5\) khi \(cos2x=-1\Rightarrow x=\dfrac{\pi}{2}+k\pi\)