Find all solutions to the trig equation 2sin(x)^2 + 3sin(x) - 2 = 0 in the range 0 <= x <= 360 degrees

Assuming knowledge of solving regular quadratic equations.Treat this as a regular quadratic equation, except it is in terms of sin(x) instead of just x as you might be used to.So using the quadratic equation (write it on the whiteboard, Toby), we get that sin(x) = ( -3 + sqrt(3^2 - 42(-2)) )/22 and sin(x) = ( -3 - sqrt(3^2 - 42*(-2)) )/2*2.These simplify down to sin(x) = 1/2 and sin(x) = -2.Multiplying through by inverse sine and remembering arcsin(sin(x)) = x, we get x = arcsin(1/2) and x=arcsin(-2).The second equation has no solution, but the first solves as x=30degrees.Bearing the sin graph in mind (draw it Toby), we see that x=1/2 crosses the graph twice in the range 0 to 360 degrees, so there is another solution we have to find.Using the cast diagram (again, draw it Toby), we know that for sine, other solutions can be found by taking our known solutions and taking them away from 180degrees. So since x =30, the other solution is 180-30=150.Total solution set for this question is therefore x=30,150 degrees