Find all solutions to the equation in the interval [0, 2).
-$\cos 2 x - \cos x = 0$
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

Question

Quizplus · Accepted Answer

Using the double-angle formula, $\cos 2x = 2\cos^2 x - 1$, we can rewrite the equation as $2\cos^2 x - 1 - \cos x = 0$, which simplifies to $2\cos^2 x - \cos x - 1 = 0$. Factoring this quadratic equation in terms of $\cos x$, we find the solutions for $\cos x$. The solutions within the given interval [0, 2π) that satisfy the original equation are $0, \frac{2\pi}{3}, \frac{4\pi}{3}$.

Find All Solutions to the Equation in the Interval [0 cos⁡2x−cos⁡x=0\cos 2 x - \cos x = 0cos2x−cosx=0

Find All Solutions to the Equation in the Interval [0 $\cos 2 x - \cos x = 0$