Solve the following equation. $2 \sin ^ { 2 } x = 6 \cos ^ { 2 } x$
A) $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
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.​ $2 \sin ^ { 2 } x = 6 \cos ^ { 2 } x$ ​
A)​ $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
B)​ $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
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

The Answer is A

Solve the Following Equation 2sin⁡2x=6cos⁡2x2 \sin ^ { 2 } x = 6 \cos ^ { 2 } x2sin2x=6cos2x

Solve the Following Equation $2 \sin ^ { 2 } x = 6 \cos ^ { 2 } x$