# Solve equation 6cos^2 2x+10cos^2x-9=0? x=?

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You should use the double angle formula for cosine such that:

`cos 2alpha = 2cos^2 alpha -1`

Hence, raising both sides to square, yields:

`cos^2 2 alpha = (2cos^2 alpha - 1)^2`

Hence, substituting `(2cos^2x - 1)^2` for `cos^2 2x` yields:

`6(2cos^2 x - 1)^2 + 10 cos^2 x - 9 = 0`

You need to expand the square using the formula `(a-b)^2= a^2 - 2ab + b^2` such that:

`6(4cos^4 x - 4cos^2 x + 1) + 10 cos^2 x - 9 = 0`

Opening the brackets yields:

`24cos^4 x - 24cos^2 x + 6 + 10 cos^2 x - 9 = 0`

`24cos^4 x - 14cos^2 x - 3 = 0`

You should come up with the following substitution `cos^2 x = y` such that:

`24y^2 - 14y - 3 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (14+-sqrt(196 + 288))/48`

`y_(1,2) = (14+-sqrt484)/48`

`y_(1,2) = (14+-22)/48 => y_1 = 36/48 => y_1 = 3/4`

`y_2 = -8/48 => y_2 = -1/6`

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

`cos^2 x = 3/4 => cos x = +-(sqrt3)/2 => x = +-pi/6 + 2npi`

`cos^2 x = -1/6` is impossible to be solved since no real squared values of x yields `-1/6` .

**Hence, evaluating the general solution to the given equation yields `x = +-pi/6 + 2npi` .**

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