# cos(x/2)+cos(x)=0 in pre calculus

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First note that `cos(2a) = 2cos^2(a) - 1` `` so `cos(x/2) = sqrt((cos(x)+1)/2)`

`cos(x/2) + cos(x) = 0`

`sqrt((cos(x)+1)/2)+cos(x) = 0`

`sqrt((cos(x)+1)/2)=-cos(x)`

`(cos(x)+1)/2 = cos^2(x)`

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

`cos(x) = (1+-sqrt(1-4(2)(-1))/4 = (1+-sqrt(1+8))/4) = (1+-3)/4`

`cos(x) = 1` or `cos(x) = -1/2`

` cos(x) = 1`

when `x = 2pi+4pin` where n is an integer.

`cos(x) = -1/2` when `x = (2pi)/3 + 4npi`

So the answer is `x = 2pi+4pin` or `x = (2pi)/3+4npi`