Solve the equation. Give a general formula for all the solutions.
-$$\sin \theta = \frac { \sqrt { 3 } } { 2 }$$
A) $\theta = \frac { \pi } { 6 } + 2 k \pi , \theta = \frac { 5 \pi } { 6 } + 2 k \pi$
B) $\theta = \frac { \pi } { 6 } + \mathrm { k } \pi , \theta = \frac { 5 \pi } { 6 } + \mathrm { k } \pi$
C) $\theta = \frac { \pi } { 3 } + 2 \mathrm { k } \pi , \theta = \frac { 2 \pi } { 3 } + 2 \mathrm { k } \pi$
D) $\theta = \frac { \pi } { 3 } + k \pi , \theta = \frac { 2 \pi } { 3 } + k \pi$

Question

Quizplus · Accepted Answer

The correct solutions are $\theta = \frac{\pi}{3} + 2k\pi$ and $\theta = \frac{2\pi}{3} + 2k\pi$, because $\sin(\theta) = \frac{\sqrt{3}}{2}$ corresponds to angles in the first and second quadrants where the sine values are positive. The angles $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ have sine values of $\frac{\sqrt{3}}{2}$, and the general solution includes adding $2k\pi$ to account for the periodic nature of the sine function, where $k$ is any integer.

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Solve the Equation $\sin \theta = \frac { \sqrt { 3 } } { 2 }$