Find the exact solutions of the given equation in the interval $[ 0,2 \pi )$ . sin 4x= -2sin 2x
A) $x = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }$
B) $x = 0 , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 }$
C) $x = \frac { \pi } { 6 } , \frac { \pi } { 2 } , \frac { 5 \pi } { 6 } , \frac { 3 \pi } { 2 }$
D) $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 }$
E) $x = \frac { 7 \pi } { 6 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 6 }$

Question

Quizplus · Accepted Answer

The equation can be rewritten using the double angle identity: sin 4x = 2sin 2x cos 2x. Thus, 2sin 2x cos 2x = -2sin 2x. Factoring out sin 2x gives sin 2x (2cos 2x + 1) = 0. This implies sin 2x = 0 or 2cos 2x + 1 = 0. Solving these gives the solutions in the interval $[0, 2\pi)$ as $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$.

Find the Exact Solutions of the Given Equation in the Interval