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

Question 95

Multiple Choice

Solve the equation in the interval [0,2π) [ 0,2 \pi ) . sinx=cos2x\sin x = \cos 2 x


A) π6,π2\frac { \pi } { 6 } , \frac { \pi } { 2 }

B) π4,π2\frac { \pi } { 4 } , \frac { \pi } { 2 }

C) π3,2π3\frac { \pi } { 3 } , \frac { 2 \pi } { 3 }

D) π6,5π6,3π2\frac { \pi } { 6 } , \frac { 5 \pi } { 6 } , \frac { 3 \pi } { 2 }

E) none

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