Solve the following equation.
$4 \sin ^ { 2 } x = 12 \cos ^ { 2 } x$
A) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B) $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
C) $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
D) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E) $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$

Question

Solve the following equation. 
$4 \sin ^ { 2 } x = 12 \cos ^ { 2 } x$ 
A) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B) $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
C) $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
D) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E) $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$

Quizplus · Accepted Answer

Using the identity $\sin^2x + \cos^2x = 1$, we can rewrite the equation as $4\sin^2x = 12(1 - \sin^2x)$. Solving for $\sin^2x$ gives $\sin^2x = \frac{3}{4}$, so $\sin x = \pm\frac{\sqrt{3}}{2}$. The solutions for $\sin x = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{3} + n\pi$ and for $\sin x = -\frac{\sqrt{3}}{2}$ are $x = \frac{2\pi}{3} + n\pi$, where $n$ is any integer.

Solve the Following Equation 4sin⁡2x=12cos⁡2x4 \sin ^ { 2 } x = 12 \cos ^ { 2 } x4sin2x=12cos2x

Solve the Following Equation $4 \sin ^ { 2 } x = 12 \cos ^ { 2 } x$