Find all solutions to the equation in the interval [0, 2).
-$\cos x = \sin 2 x$
A) $0 , \pi$
B) $\frac { \pi } { 2 } , \frac { 3 \pi } { 2 }$
C) $\frac { \pi } { 6 } , \frac { \pi } { 2 } , \frac { 5 \pi } { 6 } , \frac { 3 \pi } { 2 }$
D) $0 , \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } , \pi$

Question

Quizplus · Accepted Answer

Using the identity $\sin 2x = 2\sin x \cos x$, the equation $\cos x = \sin 2x$ can be rewritten as $\cos x = 2\sin x \cos x$. Dividing both sides by $\cos x$ (and noting that $\cos x = 0$ is not a solution in the given interval), we get $1 = 2\sin x$, which simplifies to $\sin x = \frac{1}{2}$. The solutions for $\sin x = \frac{1}{2}$ in the interval $[0, 2\pi)$ are $x = \frac{\pi}{6}, \frac{5\pi}{6}$. Additionally, since we are looking for solutions in the interval $[0, 2\pi)$, we must also consider the double angle solutions for $\sin 2x = \cos x$, which introduces additional solutions at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ due to the periodic nature of the sine and cosine functions.

Find All Solutions to the Equation in the Interval [0 cos⁡x=sin⁡2x\cos x = \sin 2 xcosx=sin2x

Find All Solutions to the Equation in the Interval [0 $\cos x = \sin 2 x$