Solve the equation on the interval $$0 \leq \theta

Question

Solve the equation on the interval $$0 \leq \theta < 2 \pi$$
-$$\sin ( 4 \theta ) = \frac { \sqrt { 3 } } { 2 }$$
A) $\frac { \pi } { 12 } , \frac { \pi } { 6 } , \frac { 2 \pi } { 3 } , \frac { 7 \pi } { 12 } , \frac { 7 \pi } { 6 } , \frac { 13 \pi } { 12 } , \frac { 5 \pi } { 3 } , \frac { 19 \pi } { 12 }$
B) 0
C) $0 , \frac { \pi } { 4 } , \pi$
D) $\frac { \pi } { 4 } , \frac { 5 \pi } { 4 }$

Quizplus · Accepted Answer

The equation $\sin(4\theta) = \frac{\sqrt{3}}{2}$ has solutions where $4\theta$ corresponds to angles in the unit circle that have a sine value of $\frac{\sqrt{3}}{2}$. These angles are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ in the first rotation (0 to $2\pi$). However, since the equation involves $4\theta$, we need to find all solutions within $0 \leq 4\theta < 8\pi$ (because multiplying the interval by 4 gives the range for $4\theta$). The solutions listed in option A correctly account for all instances of $4\theta$ that satisfy the original equation within the specified interval, considering the periodic nature of the sine function and the need to divide the found angles by 4 to solve for $\theta$.

Solve the Equation on the Interval 0≤θ<2π0 \leq \theta < 2 \pi0≤θ<2π

Solve the Equation on the Interval $0 \leq \theta < 2 \pi$