Question

Find all solutions to the equation in the interval [0, 2π).

cos 2x - cos x = 0

A.
`0, (2 \pi )/(3) , (4 \pi )/(3)`

B.
`( \pi )/(6) , (5 \pi )/(6) , (3 \pi )/(2)`

C.
`0, ( \pi )/(2), (7 \pi )/(6), (11 \pi )/(6)`

D. No solution

Answer 1

`\bf cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ \boxed{2cos^2(\theta)-1} \end{cases}\\\\ -------------------------------\\\\ cos(2x)-cos(x)=0\implies \boxed{2cos^2(x)-1}-cos(x)=0 \\\\\\ \stackrel{\textit{notice is just a quadratic}}{2cos^2(x)-cos(x)-1=0}\implies [2cos(x)+1][cos(x)-1]=0\\\\ -------------------------------`


`\bf 2cos(x)+1=0\implies 2cos(x)=-1 \\\\\\ cos(x)=-\cfrac{1}{2}\implies \measuredangle x= \begin{cases} (2\pi )/(3)\\\\ (4\pi )/(3) \end{cases}\\\\ -------------------------------\\\\ cos(x)-1=0\implies cos(x)=1\implies \measuredangle x=0`

notice that the angle at 2π also has a cosine of 1, however is out of [0, 2π).