# Solve the equation: 2cos6x-3cos4x+5=0

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You may write `cos 6x` as cos 3`*(2x) = 4cos^3 (2x) - 3 cos (2x)` and you may write `cos 4x` as cos `2*(2x) = 2cos^2 (2x) - 1` .

Substituting `4cos^3 (2x) - 3 cos(2x)` for cos 6x and `2cos^2 (2x) - 1` for cos 4x yields:

`2*(4cos^3 (2x) - 3cos(2x)) - 3(2cos^2 (2x) - 1)+ 5 = 0`

`8cos^3 (2x) - 6cos(2x) - 6cos^2 (2x) + 3 + 5 = 0`

`8cos^3 (2x) - 6cos^2 (2x) - 6cos(2x)+8 = 0`

You may form two groups such that:

`(8cos^3 (2x) - 6cos^2 (2x)) - (6cos(2x)- 8 ) = 0`

You may factor out `2cos^2 (2x)` in the first group and 2 in the second group such that:

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

You may factor out `4cos(2x) - 3` such that:

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

You may factor out 2 such that:

`2(4cos(2x) - 3)(cos^2 (2x) - 1) = 0`

`4cos(2x) - 3 = 0 =gt cos 2x = 3/4`

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

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

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

`cos 2x = 1 =gt 2x = 2npi =gt x = npi`

`cos 2x = -1 =gt 2x = +-pi + 2npi`

`x = +-pi/2 + npi`

**Hence, evaluating the complete solutions to the given equation yields `x = +-(cos^(-1)(3/4))/2 + npi ; x = npi ; x = +-pi/2 + npi.` **

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