# What is x if (cosx)^2+(cos2x)^2=1

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You should write the equation in terms of `cos x` , hence, you need to use the double angle identity, such that:

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

Squaring both sides, yields:

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

Expanding the square to the right, yields:

`(cos 2x)^2 = 4cos^4 x - 4cos^2 x + 1`

Replacing `4cos^4 x - 4cos^2 x + 1` for `(cos 2x)^2` in equation, yields:

`cos^2 x + 4cos^4 x - 4cos^2 x + 1 = 1`

`cos^2 x + 4cos^4 x - 4cos^2 x + 1 - 1 = 0`

Reducing duplicate members, yields:

`cos^2 x + 4cos^4 x - 4cos^2 x = 0`

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

Factoring out `cos^2 x` yields:

`cos^2 x(4cos^2 x - 3) = 0`

Using the zero product rule, yields:

`cos^2 x = 0 => cos x = 0 => x = +-pi/2 + 2npi`

`4cos^2 x - 3 = 0 => 4cos^2 x = 3 => cos x^2 x = 3/4`

`cos x = +-sqrt3/2 => x = +-cos^(-1)(sqrt3/2) + 2npi`

`x = +-pi/6 + 2npi`

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

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