Question 683396
Solve the equation for (x) over the interval [0(degrees),360(degrees)). What is the solution set?2sin^2(x) = sqrt(3)sin(x) 
2sin^2(x) =√3)sin(x)
2sin^2(x)-√3)sin(x=0
sinx(2sinx-√3)=0
sinx=0
x=0, 180º
or
2sinx-√3=0
sinx=√3/2
x=60º, 120º (in quadrants I and II where sin>0)