# solve 2sin^2(x)-7sin(x)=-3 for x, where 0 < x< 360degrees.

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### 1 Answer

You need to solve for x the equation `2sin^2(x)-7sin(x)=-3` , hence, you should move all terms to the left and you should come up with the notation `sin x = y` such that:

`2y^2 - 7y + 3 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (7+-sqrt(49 - 24))/4`

`y_(1,2) = (7+-sqrt(25))/4 =gt y_(1,2) = (7+-5)/4`

`y_1 = 3 ; y_2 = 1/2`

You need to solve for x the equation sin x = 3, which is a contradiction since the values of sine function are not larger than .

You need to solve for x the equation `sin x = 1/2 ` such that:

`sin x = 1/2 =gt x = sin^-1(1/2)`

You need to remember that he values of sine function are positive in 1 and 2 quadrants, hence `x=30^o ` and `x = 150^o.`

**Hence, evaluating solutions to equation in `(0^o,360^o)` yields `x=30^o` and `x = 150^o .` **