# Ineed to solve for all solutions in the domain and am having trouble with one problem: 2sin^2 x+ 3cosx- 3= 0 thanks muchthe first part in words: 2sin squared x

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### 1 Answer

You need to write the equation in terms of cos x only, hence, using the fundamental formula of trigonometry `sin^2 x + cos^2 x = 1` , you may substitute `1-cos^2 ` x for `sin^2 x ` such that:

`2(1 - cos^2 x) + 3cos x - 3 = 0`

You should come up with the following substitution such that:

`cos x = y`

`2(1 - y^2) + 3y - 3 = 0`

Opening the brackets yields:

`2 - 2y^2 + 3y - 3 = 0 => -2y^2 + 3y - 1 = 0=>2y^2- 3y + 1 = 0`

Using quadratic formula yields:

`y_(1,2) = (3+-sqrt(9 - 8))/4 => y_(1,2) = (3+-1)/4`

`y_1 = 1 ; y_2 = 1/2`

You need to solve for x the equations `cos x = y_(1,2)` such that:

`cos x = 1 => x = +-cos^(-1)(0) + 2npi`

`x = 2npi or x = +-2pi + 2npi`

`cos x = 1/2 => x = +-cos^(-1)(1/2) + 2npi`

`x = +-(pi/3) + 2npi`

**Hence, evaluating the general solutions to the given trigonometric equation yields `x = 2npi, x = +-2pi + 2npi, x = +-(pi/3) + 2npi.` **

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