# in the interval 0 degrees <= A < 360 degrees, how many values of A satisfy the equation sin^3A- 2sin^2A= 3sinA

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### 1 Answer

You need to move all terms containing `sin A` to the left side such that:

`sin^3 A - 2sin^2 A - 3sin A = 0`

You need to factor out sin A such that:

`sin A*(sin^2A - 2sin A - 3) = 0`

`sin A = 0 =gt A = sin^(-1) 0`

`A = 0^o`

`A = pi = 180^o`

You need to solve the equation `sin^2A - 2sin A - 3 = 0`

You should come up with the substitution `sin A = x` , hence you may write the equation in terms of x such that:

`x^2 - 2x - 3 = 0`

You should use the quadratic formula to find `x_1` and `x_2` such that:

`x_(1,2) = (2+-sqrt(4+12))/2 =gt x_(1,2) = (2+-4)/2`

`x_1 = 3 ; x_2 = -1`

You need to find the angle A, hence you need to solve the equations `sin A = 3` and `sin A = -1`

`sin A = 3` contradiction since sin A is not larger than 1.

`sin A = -1 =gt A = 3pi/2 = 270^o`

**Hence, evaluating solutions to equation `sin^3 A - 2sin^2 A - 3sin A = 0 ` , `A in [0^o,360^o)` , yields `A = 0^o, A = 180^o,A = 270^o .` **