Question

Given that 2cos^2 x - 3sin x -3 = 0, show that 2sin^2 x + 3sin x +1 = 0?

Hence solve 2cos^2 x + 4sin x - 3 = 0, giving all solutions in the interval 0 ≤ x ≤ 2π

Answer 1

2 cos^2 x - 3 sin x - 3 = 0
but
cos^2 x = 1 - sin^2 x
so
2 (1 -sin^2 x) - 3 sin 3 - 3 = 0
2 - 2 sin^2 x - 3 sin x - 3 = 0
2 sin^2 x + 3 sin x + 1 = 0
(2 sin x + 1)(sin x + 1) = 0
sin x = -1/2 or sin x = -1

x = 210 degrees or x = 270 degrees

now do the second half :)

Answer 2

oh, forgot quadrant 4
360 - 30 = 330 degrees