Solve for the angle , where 0 2
-$$\sin ^ { 2 } \theta = \frac { 1 } { 4 }$$
A) $\theta = \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } , \frac { 7 \pi } { 6 } , \frac { 11 \pi } { 6 }$
B) $\theta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }$
C) $\theta = 0 , \pi , 2 \pi$
D) $\theta = \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 } , \frac { 5 \pi } { 3 }$

Question

Quizplus · Accepted Answer

Taking square roots of both sides, we get $\sin 	heta = \pm \frac{1}{2}$. The angles where $\sin 	heta = \frac{1}{2}$ are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$, and the angles where $\sin	heta = -\frac{1}{2}$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. These are all in the interval $(0,2\pi)$, so the answer is A).

Solve for the Angle , Where 0 2 - sin⁡2θ=14\sin ^ { 2 } \theta = \frac { 1 } { 4 }sin2θ=41​

Solve for the Angle , Where 0 2
- $\sin ^ { 2 } \theta = \frac { 1 } { 4 }$