Answer :
we have the equation
[tex]2cos^2x=cos\left(x\right)[/tex]Simplify
[tex]\begin{gathered} \frac{2cos^2x}{cos(x)}=\frac{cos\left(x\right)}{xos(x)} \\ 2cos(x)=1 \\ cos(x)=\frac{1}{2} \end{gathered}[/tex]Remember that
[tex]cos(\frac{pi}{3})=\frac{1}{2}[/tex]the value of the cosine is positive
that means
the angle x lies on the first quadrant and IV quadrant
For the First quadrant x=pi/3 radians
For the IV quadrant
x=2pi-pi/3
x=5pi/3 radians
therefore
In the given interval for x
the solutions are