ĐKXĐ: cosx>0
=>\(x\in\left(\dfrac{3}{2}\Omega+k2\Omega;\dfrac{5}{2}\Omega+k2\Omega\right)\cup\left(-\dfrac{\Omega}{2}+k2\Omega;\dfrac{\Omega}{2}+k2\Omega\right)\)
\(\dfrac{cos2x-cosx}{\sqrt{cosx}}=0\)
=>\(cos2x-cosx=0\)
=>\(cos2x=cosx\)
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được: \(x=k2\Omega\)