\(\Leftrightarrow\left(-2sin3x.sinx\right)^2=5+sin3x\)
\(\Leftrightarrow4sin^23x.sin^2x=5+sin3x\)
Do \(\left\{{}\begin{matrix}sin^23x\le1\\sin^2x\le1\end{matrix}\right.\) \(\Rightarrow4sin^23x.sin^2x\le4\)
\(sin3x\ge-1\Rightarrow5+sin3x\ge4\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin^2x=1\\sin3x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(sinx-1\right)\left(sinx+1\right)=1\\4sin^3x-3sinx-1=0\end{matrix}\right.\)
\(\Leftrightarrow sinx=1\)
\(\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)
\(\left(cos4x-cos2x\right)^2=5+2sinx\)
\(\Leftrightarrow4sin^23x.sin^2x=5+2sinx\)
Ta có: \(sinx\in\left[-1;1\right];sin3x\in\left[-1;1\right]\)
\(\Rightarrow\left\{{}\begin{matrix}4sin^23x.sin^2x\ge4\\5+sin3x\le4\end{matrix}\right.\)
\(\Rightarrow4sin^23x.sin^2x\ge5+sin3x\)
Đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}sin^23x=1\\sin^2x=1\\sin3x=-1\end{matrix}\right.\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
