Find all solutions in the interval [0, 2π).
-$$\sin \left( \frac { x } { 2 } \right) = 2 \cos ^ { 2 } x - 1$$
A) $\frac { \pi } { 5 } , \pi , \frac { 9 \pi } { 5 }$
B) $\frac { \pi } { 3 } , \pi , \frac { 5 \pi } { 3 }$
C) $\frac { \pi } { 5 } , \frac { 9 \pi } { 5 }$
D) $0 , \pi$

Question

Quizplus · Accepted Answer

Using the double-angle formula for cosine, we have:$$\sin \left( \frac { x } { 2 } \right) = 2 \cos ^ { 2 } x - 1$$$$= 2 (1 - \sin ^ { 2 } x) - 1$$$$= 1 - 2 \sin ^ { 2 } x$$$$= \cos 2x$$Therefore, the solutions are:$$2x = \pm \frac { \pi } { 2 } + 2k\pi$$$$x = \pm \frac { \pi } { 4 } + k\pi$$In the interval [0, 2π), the solutions are:$$x = \frac { \pi } { 4 }, \pi , \frac { 9 \pi } { 4 }$$

Find All Solutions in the Interval [0, 2π) sin⁡(x2)=2cos⁡2x−1\sin \left( \frac { x } { 2 } \right) = 2 \cos ^ { 2 } x - 1sin(2x​)=2cos2x−1

Find All Solutions in the Interval [0, 2π) $\sin \left( \frac { x } { 2 } \right) = 2 \cos ^ { 2 } x - 1$