# Determine the general solution of : `cos^2 theta + 3sin theta = - 3` Thank you

*print*Print*list*Cite

You need to use the basic trigonometric formula `1 - sin^2 theta = cos^2 theta` , to write the equation in terms of `sin theta` , only, such that:

`(1 - sin^2 theta) + 3sin theta = -3`

`1 - sin^2 theta + 3sin theta + 3 = 0`

`- sin^2 theta + 3sin theta +4 = 0`

Factoring out -1 yields:

`-1(sin^2 theta - 3sin theta - 4) = 0`

Replacing t for `sin theta` yields:

`t^2 - 3t - 4 = 0`

You may use quadratic formula to evaluate `t_(1,2)` such that:

`t_(1,2) = (3+-sqrt(9 + 16))/2 => t_(1,2) = (3+-5)/2`

`t_1 = 4, t_2 = -1`

Replacing back sin theta for t yields:

`sin theta = 4` invalid since `sin theta <= 1`

`sin theta = -1 => theta = (-1)^(n+1)(pi/2) + n*pi`

**Hence, evaluating the general solution to the given equation, yields **`theta = (-1)^(n+1)(pi/2) + n*pi.`