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 = 2 \sin 2x \cos 2x $. Thus, $ 2 \sin 2x \cos 2x = -2 \sin 2x $. Factoring out $ \sin 2x $, we get $ \sin 2x (2 \cos 2x + 1) = 0 $. This gives solutions $ \sin 2x = 0 $ or $ 2 \cos 2x + 1 = 0 $. Solving these, we find $ x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} $.

Find the Exact Solutions of the Given Equation in the Interval