Question

Giải phương trình

\(\left(\sqrt{1-cosx}+\sqrt{cosx}\right)cos2x=\frac{1}{2}sin4x\)

Answer 1

ĐKXĐ: \(cosx\ge0\)

\(\Leftrightarrow\left(\sqrt{1-cosx}+\sqrt{cosx}\right)cos2x=sin2x.cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\left(1\right)\\\sqrt{1-cosx}+\sqrt{cosx}=sin2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2cos^2x-1=0\)

\(\Leftrightarrow cos^2x=\frac{1}{2}\Rightarrow cosx=\frac{\sqrt{2}}{2}\)

\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

Xét (2): ta có \(VP=sin2x\le1\)

\(VT=\sqrt{1-cosx}+\sqrt{cosx}\ge\sqrt{1-cosx+cosx}=1\ge VP\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}cosx=0\\sin2x=1\end{matrix}\right.\\\left\{{}\begin{matrix}cosx=1\\sin2x=1\end{matrix}\right.\end{matrix}\right.\) (đều vô nghiệm)