Solve the equation on the interval $$0 \leq \theta

Question

Solve the equation on the interval $$0 \leq \theta < 2 \pi$$
-$$2 \cos ( 2 \theta ) = \sqrt { 3 }$$
A) $\frac { \pi } { 2 }$
B) $\frac { 3 \pi } { 2 }$
C) $\frac { \pi } { 6 } , \frac { 11 \pi } { 6 }$
D) $\frac { \pi } { 12 } , \frac { 11 \pi } { 12 } , \frac { 13 \pi } { 12 } , \frac { 23 \pi } { 12 }$

Quizplus · Accepted Answer

First, solve for $\cos(2\theta) = \frac{\sqrt{3}}{2}$. The solutions for $\cos(x) = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$ within one cycle of the cosine function. Since we're solving for $2\theta$, we set $2\theta = \frac{\pi}{6}$ and $2\theta = \frac{11\pi}{6}$, giving $\theta = \frac{\pi}{12}$, $\theta = \frac{11\pi}{12}$, $\theta = \frac{13\pi}{12}$, and $\theta = \frac{23\pi}{12}$, which are all within the given interval $0 \leq \theta < 2\pi$.

Solve the Equation on the Interval 0≤θ<2π0 \leq \theta < 2 \pi0≤θ<2π

Solve the Equation on the Interval $0 \leq \theta < 2 \pi$