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

Question

Solve the equation on the interval $$0 \leq \theta < 2 \pi \text {. }$$
-$\sin \theta + \sqrt { 3 } \cos \theta = - 1$
A) $\frac { 3 \pi } { 2 } , \frac { 5 \pi } { 6 }$

B) $0 , \frac { 2 \pi } { 3 }$

C) $\frac { \pi } { 2 } , \frac { 7 \pi } { 6 }$

D) $\frac { 3 \pi } { 2 } , \frac { \pi } { 6 }$

Quizplus · Accepted Answer

Divide the equation by 2 and use the identity $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha$, with $\alpha = \frac{\pi}{3}$, to rewrite the equation as $\sin(\theta - \frac{\pi}{3}) = -\frac{1}{2}$. Solving for $\theta$, we find $\theta = \frac{5\pi}{6}$ and $\theta = \frac{3\pi}{2}$, which are the solutions within the given interval.

Solve the Equation on the Interval 0≤θ<2π. 0 \leq \theta < 2 \pi \text {. }0≤θ<2π.

Solve the Equation on the Interval $0 \leq \theta < 2 \pi \text {. }$