SOLUTION: For what values of t on the interval [0,2pi]is sin(t)= Square root of -3/2

Question 867443: For what values of t on the interval [0,2pi]is sin(t)= Square root of -3/2
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
For what values of t on the interval [0,2pi]is sin(t)= Square root of -3/2
Note: I believe you meant the minus sign to be in front of the sqrt sign, like: -√3/2
You can't take the sqrt of a negative number.
sin(t)= -√3/2
t=4π/3, 5π/3 (In quadrants III and IV in which sin<0)

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