Given the equation below, determine two solutions such that $0 ^ { \circ } \leq \theta

Question

Given the equation below, determine two solutions such that $0 ^ { \circ } \leq \theta < 360 ^ { \circ }$.
$$\sin \theta = - \frac { \sqrt { 3 } } { 2 }$$
A) $\theta = 120 ^ { \circ } , 240 ^ { \circ }$
B) $\theta = 240 ^ { \circ } , 300 ^ { \circ }$
C) $\theta = 150 ^ { \circ } , 210 ^ { \circ }$
D) $\theta = 135 ^ { \circ } , 225 ^ { \circ }$
E) $\theta = 225 ^ { \circ } , 315 ^ { \circ }$

Quizplus · Accepted Answer

The equation $\sin \theta = - \frac { \sqrt { 3 } } { 2 }$ indicates that $\theta$ is in the third or fourth quadrant (since sine is negative in these quadrants) and corresponds to an angle whose reference angle is $60^\circ$. The solutions within the given range are $240^\circ$ (in the third quadrant) and $300^\circ$ (in the fourth quadrant).

Given the Equation Below, Determine Two Solutions Such That 0∘≤θ<360∘0 ^ { \circ } \leq \theta < 360 ^ { \circ }0∘≤θ<360∘

Given the Equation Below, Determine Two Solutions Such That $0 ^ { \circ } \leq \theta < 360 ^ { \circ }$