Hàm số \(y=2cot\left(\dfrac{x}{3}+\dfrac{\pi}{4}\right)\) tuần hoàn với chu kì \(T=\dfrac{\pi}{\left|\dfrac{1}{3}\right|}=3\pi\).

Bạn có thể giải chi tiết đc k ạ 

\(sin\in\left[-1;1\right]\Rightarrow y=3-4sinx\in\left[-1;7\right]\)

\(\Rightarrow y_{min}=-1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=7\Leftrightarrow sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(sin2x-cos3x=0\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{2}\right)-cos3x=0\)

\(\Leftrightarrow-2sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right).sin\left(-\dfrac{x}{2}-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right)=0\\sin\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5x}{2}-\dfrac{\pi}{4}=k\pi\\\dfrac{x}{2}+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(y=2sin^2x-cos2x=-\left(1-2sin^2x\right)-cos2x+1=1-2cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=1-2cos2x\in\left[-1;3\right]\)

\(\Rightarrow y_{min}=-1\Leftrightarrow cos2x=1\Leftrightarrow x=k\pi\)

\(y_{max}=3\Leftrightarrow cos2x=-1\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

Ta có y= 2sin2 x+ cos2 2x = 2sin2 x + (1- 2sin2x)2

\(sin\left(2x+\dfrac{\pi}{4}\right)+cosx=0\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{4}\right)+cosx=0\)

\(\Leftrightarrow2cos\left(\dfrac{3x}{2}-\dfrac{\pi}{8}\right).cos\left(\dfrac{x}{2}-\dfrac{\pi}{8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{3x}{2}-\dfrac{\pi}{8}\right)=0\\cos\left(\dfrac{x}{2}-\dfrac{\pi}{8}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-\dfrac{\pi}{8}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}-\dfrac{\pi}{8}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)