Solve the equation in the interval $[ 0,2 \pi )$ . $\sin x = \cos 2 x$
A) $\frac { \pi } { 6 } , \frac { \pi } { 2 }$

B) $\frac { \pi } { 4 } , \frac { \pi } { 2 }$

C) $\frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$

D) $\frac { \pi } { 6 } , \frac { 5 \pi } { 6 } , \frac { 3 \pi } { 2 }$

E) none

Question

Solve the equation in the interval $[ 0,2 \pi )$ . $\sin x = \cos 2 x$
A) $\frac { \pi } { 6 } , \frac { \pi } { 2 }$

B) $\frac { \pi } { 4 } , \frac { \pi } { 2 }$

C) $\frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$

D) $\frac { \pi } { 6 } , \frac { 5 \pi } { 6 } , \frac { 3 \pi } { 2 }$

E) none

Quizplus · Accepted Answer

Using the identity $\cos 2x=1-2\sin^2 x,$ we have $\sin x=1-2\sin^2 x$ or $2\sin^2 x+\sin x-1=0.$ Factoring yields $(2\sin x-1)(\sin x+1)=0.$ Thus $\sin x=-1$ or $\sin x=\frac12.$ Since $-1$ is not in the interval $[0,2\pi),$ we solve $\sin x=\frac12.$ The solutions in $[0,2\pi)$ are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}.$ Finally, $\sin \frac{3\pi}{2}=-1,$ so that completes our solutions: $\frac{\pi}{6}, \frac{5\pi}{6},$ and $\frac{3\pi}{2}.$

Solve the Equation in the Interval [0,2π)[ 0,2 \pi )[0,2π) sin⁡x=cos⁡2x\sin x = \cos 2 xsinx=cos2x

Solve the Equation in the Interval $[ 0,2 \pi )$ $\sin x = \cos 2 x$