Question 867443
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=4&#960;/3, 5&#960;/3 (In quadrants III and IV in which sin<0)