SOLUTION: Find all real numbers in [0, 2pi] that satisfy the equation. sin4x=Square root 3/2

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Question 966629: Find all real numbers in [0, 2pi] that satisfy the equation.
sin4x=Square root 3/2
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find all real numbers in [0, 2pi] that satisfy the equation.
sin4x=Square root 3/2
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sin(4x)=√3/2
4x=pi/3, 2pi/3
x=pi/12, pi/6