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Solve the Equation on the Interval [0, 2π) sin4x=32\sin 4 x = \frac { \sqrt { 3 } } { 2 }

Question 96

Multiple Choice

Solve the equation on the interval [0, 2π) .
- sin4x=32\sin 4 x = \frac { \sqrt { 3 } } { 2 }


A) π12,π6,2π3,7π12,7π6,13π12,5π3,19π12\frac { \pi } { 12 } , \frac { \pi } { 6 } , \frac { 2 \pi } { 3 } , \frac { 7 \pi } { 12 } , \frac { 7 \pi } { 6 } , \frac { 13 \pi } { 12 } , \frac { 5 \pi } { 3 } , \frac { 19 \pi } { 12 }
B) π4,5π4\frac { \pi } { 4 } , \frac { 5 \pi } { 4 }
C) 0,π4,π0 , \frac { \pi } { 4 } , \pi
D) 0

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