# What is x in cos3x+cos7x=cos2x? show all the work

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You need to transform the sum to the left into a product such that:

`cos3x+cos7x = 2cos((3x+7x)/2)cos((3x-7x)/2)`

`cos3x+cos7x = 2cos(10x/2)cos(-4x/2)`

Since the cosine function is even, hence `cos(-4x/2) = cos(4x/2)`

`cos3x+cos7x = 2cos(5x)cos(2x)`

Hence, substituting `2cos(5x)cos(2x)` for `cos3x+cos7x` in equation yields:

`2cos(5x)cos(2x) = cos(2x)`

Moving all terms to one side yields:

`2cos(5x)cos(2x)- cos(2x) = 0`

Factoring out `cos(2x)` yields:

` cos(2x)(2cos(5x) - 1) = 0`

`cos(2x) = 0 => 2x = +-pi/2 + 2npi => x = +-pi/4 + npi`

`2cos(5x) - 1 = 0 => 2cos(5x) = 1 => cos(5x) = 1/2 => 5x = +-pi/3 + 2npi`

`x = +-pi/15 + 2npi/5`

**Hence, evaluating the general solutions to the given equation yields `x = +-pi/4 + npi` and `x = +-pi/15 + 2npi/5` .**