# Solve for theta: `2*cos^2 theta + 7*sin theta -5= 0`, `0<= theta<= 720` 0

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### 1 Answer

The equation `2cos^2 theta + 7 sin theta -5 = 0` has to be solved for `0 <= theta <= 720`

`2cos^2 theta + 7 sin theta -5 = 0`

=> `2(1 - sin^2 theta) + 7 sin theta -5 = 0`

=> `2 - 2*sin^2 theta + 7 sin theta -5 = 0`

=> `- 2*sin^2 theta + 7 sin theta -3 = 0`

=> `2*sin^2 theta - 7 sin theta +3 = 0`

=> `2*sin^2 theta - 6 sin theta - sin theta +3 = 0`

=> `2*sin theta*(sin theta - 3) - 1(sin theta - 3) = 0`

=> `(2*sin theta - 1)*(sin theta - 3) = 0`

`sin theta = 1/2` and `sin theta = 3`

Ignore the root `sin theta = 3` as `sin theta` lies in [-1, 1]

`sin theta = 1/2`

`theta = 30, 150, 390, 510`

**The solution of `theta` in the given domain is **`theta = 30, 150, 390, 510`