Question

I need help with this practice problem It has two additional pictures of the answer options that go along, I will send you these

Answer 1

Given the trigonometric equation:

`2\cos ^2x=\cos x`

We will solve the equation as follows:

`\begin{gathered} 2\cos ^2x-\cos x=0 \\ \cos x(2\cos x-1)=0 \\ \cos x=0\rightarrow x=(\pi)/(2),(3\pi)/(2) \\ 2\cos x=1\rightarrow\cos x=(1)/(2)\rightarrow x=(\pi)/(3),(5\pi)/(3) \end{gathered}`

So, the answer will be:

`x=\mleft\lbrace(\pi)/(3),(\pi)/(2),(3\pi)/(2),(5\pi)/(3)\mright\rbrace`