Question

How do you solve `2 cos^2 x = 3 sin x` on the interval `[0,2pi]`?

Answer 1

Solve `2cos^2 x = 3sin x`

Ans: `pi/6` and `(5pi)/6`

Explanation:

`2(1 - sin^2 x) = 3sin x.` Call sin x = t
`2 - 2t^2 = 3t`--> `2t^2 + 3t - 2 = 0`
D = d^2 = 9 + 16 = 25 --> d = +- 5
`t = sin x = -3/4 +- 5/4`

`t = 2/4 = 1/2` and `t = -8/4 = - 2` (rejected because < -1)

`t = sin x = 1/2` --> `x = pi/6` and `x = (5pi)/6`