Question

Solve the following equation for all values of x, 0 ≤ x ≤ 2π2cos^2 (x) = cos(x)

Answer 1

we have the equation

$2cos^2x=cos\left(x\right)$

Simplify

$\begin{gathered} \frac{2cos^2x}{cos(x)}=\frac{cos\left(x\right)}{xos(x)} \\ 2cos(x)=1 \\ cos(x)=\frac{1}{2} \end{gathered}$

Remember that

$cos(\frac{pi}{3})=\frac{1}{2}$

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

x=pi/3 and 5pi/3