Answer
$\frac{2\pi}{3},\frac{4\pi}{3}$
Work Step by Step
1. $2cos^2\theta-7cos\theta-4=0 \Longrightarrow (2cos\theta+1)(cos\theta-4)=0 \Longrightarrow cos\theta=-\frac{1}{2}, cos\theta=4$,
2. $cos\theta=4 \Longrightarrow $ no solution in $[0,2\pi)$,
3. $cos\theta=-\frac{1}{2} \Longrightarrow \theta=\frac{2\pi}{3},\frac{4\pi}{3}$ in $[0,2\pi)$.
