# How to solve equation if 0<x <2pi cos^22x+cos^24x=1?

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### 1 Answer

You should use the cosine of double angle identity such that:

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

You need to raise to square cos 4x such that:

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

You should expand the square such that:

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

You need to substitute `4cos^4 2x - 4cos^2 2x + 1` for `cos^2 4x` such that:

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

Reducing like terms yields:

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

You need to factor out `cos^2 2x` such that:

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

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

`2x = pi/2or 2x = 3pi/2`

`x = pi/4 ; x = 3pi/4 `

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

`cos 2x = +-sqrt3/2`

`2x = pi/6 or 2x = 2pi - pi/6`

`x = pi/12 or x = 11pi/12`

`2x = 5pi/6 or 2x = 7pi/6`

`x = 5pi/12 or x = 7pi/12`

**Hence, evaluating the solutions to equation in `(0,2pi)` yields `pi/12 ; pi/4 ; 3pi/4 ; 5pi/12 ; 7pi/12 ; 11pi/12` .**

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