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\(y=\left|2sin^2x-sinx-1\right|-2sinx\)
Đặt \(sinx=t\in\left[-1;1\right]\)
\(\Rightarrow y=f\left(t\right)=\left|2t^2-t-1\right|-2t\)
BBT cho \(f\left(t\right)\) trên \(\left[-1;1\right]\):
Từ BBT ta thấy \(y_{max}=4\) khi \(sinx=-1\); \(y_{min}=-2\) khi \(sinx=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\)
\(y=\left|sinx+cos2x\right|=\left|2sin^2x-sinx-1\right|\)
\(\Leftrightarrow y=\left|f\left(t\right)\right|=\left|2t^2-t-1\right|\)
\(f\left(-1\right)=2\Rightarrow y=2\)
\(f\left(1\right)=0\Rightarrow y=0\)
\(f\left(\dfrac{1}{4}\right)=-\dfrac{9}{8}\Rightarrow y=\dfrac{9}{8}\)
\(\Rightarrow y_{min}=0;y_{max}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=\sqrt{3}cos2x+2sinxcosx-2\)
\(=\sqrt{3}cos2x+sin2x-2\)
Ta có: \(\left|\sqrt{3}cos2x+sin2x\right|\le\sqrt{\left(\sqrt{3}\right)^2+1^2}=2\)
Do đó \(-2\le\sqrt{3}cos2x+sin2x\le2\)
\(\Leftrightarrow-4\le\sqrt{3}cos2x+sin2x-2\le2\).
Ta có: \(\left|\sqrt{3}cosx-sinx\right|\le\sqrt{\left(\sqrt{3}\right)^2+\left(-1\right)^2}=2\)
Do đó \(-2\le\sqrt{3}cosx-sinx\le2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=1-2\sin^2x-\sin x+3=-2\sin^2x-\sin x+4\)
\(\sin x=t;t\in\left[-1;1\right]\)
Xét hàm f(t) trên [-1;1]
\(f\left(-1\right)=-2+1+4=3\)
\(f\left(1\right)=-2-1+4=1\)
\(f\left(-\frac{1}{4}\right)=-2.\frac{1}{16}+\frac{1}{4}+4=\frac{33}{8}\)
\(\Rightarrow\left\{{}\begin{matrix}y_{max}=\frac{33}{8};"="\Leftrightarrow\sin x=-\frac{1}{4}\Rightarrow x=...\\y_{min}=1;"="\Leftrightarrow\sin x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
\(y=\left|sinx-\left(1-2sin^2x\right)\right|=\left|2sin^2x+sinx-1\right|\)
Đặt \(t=sinx;-1\le t\le1\)
\(\Rightarrow y=\left|2t^2+t-1\right|\)
Đặt \(f\left(t\right)=2t^2+t-1;-1\le t\le1\)
Vẽ BBT của \(f\left(t\right)=2t^2+t-1;-1\le t\le1\) sẽ tìm được \(f\left(t\right)_{min}=-\dfrac{9}{8};f\left(t\right)_{max}=2\)
\(\Rightarrow0\le\left|f\left(t\right)\right|\le2\)
\(\Leftrightarrow0\le y\le2\)
\(\Rightarrow y_{min}=0\Leftrightarrow2sin^2x+sinx-1=0\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(y_{max}=2\Leftrightarrow t=1\Leftrightarrow sinx=1\)