ĐKXĐ: \(cosx\ge0\)
\(\Leftrightarrow\left(\sqrt{1-cosx}+\sqrt{cosx}\right)cos2x=sin2x.cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\sqrt{1-cosx}+\sqrt{cosx}=sin2x\end{matrix}\right.\)
TH1: \(cos2x=0\Rightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
Kết hợp điều kiện \(cosx\ge0\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
TH2: \(\sqrt{1-cosx}+\sqrt{cosx}=sin2x\)
Ta có \(VT=\sqrt{1-cosx}+\sqrt{cosx}\ge\sqrt{1-cosx+cosx}=1\)
\(VP=sin2x\le1\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}sin2x=1\\\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow ptvn\)
