# Solve for x equation 2sin^2x/2+cos2x-1=0?? is hard trigo eq

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You should use half angle and double angle identities such that:

`2sin^2(x/2) = 1 - cos (2*(x/2)) => 2sin^2(x/2) = 1 - cos x`

`cos 2x = 2cos^2 x - 1`

Hence, substituting `1 - cos x` for `2sin^2(x/2)` and `2cos^2 x - 1` for `cos 2x ` such that:

`1 - cos x + 2cos^2 x - 1 - 1 = 0`

`2cos^2 x - cos x - 1 = 0`

You should come up with the substitution `cos x = y` such that:

`2y^2 - y - 1 = 0`

Using quadratic formula yields:

`y_(1,2) = (1+-sqrt(1+8))/4 => y_(1,2) = (1+-sqrt9)/4`

`y_(1,2) = (1+-3)/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)(1) + 2npi => x = 0+2npi => x=2npi`

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

`x = pi - pi/3 + 2npi => x = 2pi/3 + 2npi`

**Hence, evaluating the solutions to the given equation yields `x=2npi` and `x = 2pi/3 + 2npi` .**

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