Solve the following equation. $\cos 2 x ( 4 \cos x + 2 ) = 0$
A) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi , \frac { 4 \pi } { 3 } + 2 n \pi$
B) $\frac { \pi } { 4 } + \frac { n \pi } { 2 } , \frac { 2 \pi } { 3 } + 2 n \pi , \frac { 4 \pi } { 3 } + 2 n \pi$
C) $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
D) $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
E) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$

Question

Solve the following equation.   $\cos 2 x ( 4 \cos x + 2 ) = 0$  
A) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi , \frac { 4 \pi } { 3 } + 2 n \pi$
B) $\frac { \pi } { 4 } + \frac { n \pi } { 2 } , \frac { 2 \pi } { 3 } + 2 n \pi , \frac { 4 \pi } { 3 } + 2 n \pi$
C)  $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
D)  $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
E)  $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$

Quizplus · Accepted Answer

The equation $\cos 2x (4 \cos x + 2) = 0$ can be solved by setting each factor equal to zero. For $\cos 2x = 0$, the solutions are $x = \frac{\pi}{4} + \frac{n\pi}{2}$. For $4\cos x + 2 = 0$, solving for $\cos x = -\frac{1}{2}$ gives $x = \frac{2\pi}{3} + 2n\pi$ and $x = \frac{4\pi}{3} + 2n\pi$.

Solve the Following Equation cos⁡2x(4cos⁡x+2)=0\cos 2 x ( 4 \cos x + 2 ) = 0cos2x(4cosx+2)=0

Solve the Following Equation $\cos 2 x ( 4 \cos x + 2 ) = 0$