# Solve for x: cos (3x) = - cos x

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Solve cos(3x)=-cosx:

cos(3x)+cosx=0

Use a trigonometric identity to rewrite cos(3x):

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

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

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

(a) ` `2cos(x)=0 ==> cos(x)=0 ==> `x=pi/2+n pi, n in ZZ`

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

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

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

`x=pi/4 + (npi)/2`

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The solutions are `x=pi/2 + n pi, pi/4 + (n pi)/2`

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The graph of cos(3x)+cos(x) (solutions are zeros):

or the graphs of cos(3x) in black, -cos(x) in red -- the solutions are the intersections:

** If you don't know the identity to get `cos(3x)=4cos^3(x)-3cos(x)` you can use the sum formula:

cos(3x)=cos(2x+x)=cos(2x)cos(x)-sin(2x)sin(x)

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

`=2cos^3(x)-cos(x)-2[(1-cos^2(x))cos(x)]`

`=2cos^3(x)-cos(x)-2cos(x)+2cos^3(x)`

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

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What is the identity for cos3x

We have given

`cos(3x)=-cos(x)`

But

`cos(3x)=4cos^3(x)-3cos(x)` ,so problem reduces to

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

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

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

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

`either`

`2cos(x)=0`

`or`

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

`If`

`cos(x)=0`

`` Then `x=2npi+-pi/2` , n is an integer.

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

Then

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

**If**

`cos(x)=cos(pi/4)`

`x=2n_1pi+-pi/4 ,` `n_1` is an integer .

**If**

`cos(x)=-cos(pi/4)=cos(pi+pi/4)`

`=cos((5pi)/4)`

`x=2n_2pi+-(5pi)/4` , `n_2` is an integer.