Solve the following equation. $8 \sin ^ { 2 } 2 x = 4$
A) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B) $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
C) $\frac { \pi } { 8 } + \frac { n \pi } { 2 } , \frac { 3 \pi } { 8 } + \frac { n \pi } { 2 }$
D) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E) $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$

Question

Solve the following equation.​ $8 \sin ^ { 2 } 2 x = 4$ ​
A) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B)​ $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
C)​ $\frac { \pi } { 8 } + \frac { n \pi } { 2 } , \frac { 3 \pi } { 8 } + \frac { n \pi } { 2 }$
D)​ $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E)​ $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$

Quizplus · Accepted Answer

First, simplify the equation: $8 \sin^2 2x = 4$ becomes $\sin^2 2x = \frac{1}{2}$. Taking the square root gives $\sin 2x = \pm \frac{\sqrt{2}}{2}$, which corresponds to angles of $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ within one cycle. Since the equation involves $2x$, we divide these angles by 2 to solve for $x$, resulting in $x = \frac{\pi}{8}$ and $x = \frac{3\pi}{8}$. The general solution for a sine function includes all possible cycles, hence the $+\frac{n\pi}{2}$ to account for the period of the sine function after considering the $2x$ factor.

Solve the Following Equation 8sin⁡22x=48 \sin ^ { 2 } 2 x = 48sin22x=4

Solve the Following Equation $8 \sin ^ { 2 } 2 x = 4$