Determine all solutions of the equation in radians.
-$\cos 2 \theta = \frac { 5 } { 6 }$ and $\theta$ terminates in quadrant $\mathrm { I } \quad$ Find $\sin \theta$.
A) $\sin \theta = \frac { \sqrt { 3 } } { 6 }$
B) $\sin \theta = 0$
C) $\sin \theta = 2 \sqrt { 3 }$
D) $\sin \theta = - \frac { \sqrt { 3 } } { 6 }$

Question

Quizplus · Accepted Answer

Given $\cos 2\theta = \frac{5}{6}$, we use the double angle identity $\cos 2\theta = 2\cos^2\theta - 1$ to find $\cos\theta$. Solving for $\cos\theta$, we get $\cos\theta = \sqrt{\frac{11}{12}}$. Since $\theta$ is in the first quadrant, where sine values are positive, we use $\sin^2\theta + \cos^2\theta = 1$ to find $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left(\frac{11}{12}\right)} = \sqrt{\frac{1}{12}} = \frac{\sqrt{3}}{6}$.

Determine All Solutions of the Equation in Radians cos⁡2θ=56\cos 2 \theta = \frac { 5 } { 6 }cos2θ=65​

Determine All Solutions of the Equation in Radians $\cos 2 \theta = \frac { 5 } { 6 }$