Given the trigonometric equation:
[tex]2\cos ^2x=\cos x[/tex]
We will solve the equation as follows:
[tex]\begin{gathered} 2\cos ^2x-\cos x=0 \\ \cos x(2\cos x-1)=0 \\ \cos x=0\rightarrow x=\frac{\pi}{2},\frac{3\pi}{2} \\ 2\cos x=1\rightarrow\cos x=\frac{1}{2}\rightarrow x=\frac{\pi}{3},\frac{5\pi}{3} \end{gathered}[/tex]
So, the answer will be:
[tex]x=\mleft\lbrace\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{3}\mright\rbrace[/tex]
