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) $\theta = \frac { \pi } { 3 } + 2 k \pi , \theta = \frac { 2 \pi } { 3 } + 2 k \pi$
D) $\theta = \frac { \pi } { 3 } + \mathrm { k } \pi , \theta = \frac { 2 \pi } { 3 } + \mathrm { k } \pi$

Question

Quizplus · Accepted Answer

The correct values of $\theta$ that satisfy $\sin \theta = \frac{\sqrt{3}}{2}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, as these are the angles in the first and second quadrants where the sine value is $\frac{\sqrt{3}}{2}$. The general solution includes these angles plus any integer multiple of $2\pi$ to account for the periodic nature of the sine function.

Solve the Equation sin⁡θ=32\sin \theta = \frac { \sqrt { 3 } } { 2 }sinθ=23​​

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