Question

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

cos 2x - cos x = 0

A. $0, \frac{2 \pi }{3} , \frac{4 \pi }{3}$

B. $\frac{ \pi }{6} , \frac{5 \pi }{6} , \frac{3 \pi }{2}$

C.$0, \frac{ \pi }{2}, \frac{7 \pi }{6}, \frac{11 \pi }{6}$

D. No solution

Answer 1

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

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

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