# solve cos^4x-1=0 for all values of x any help is greatly appreciated(: `cos^4x-1=0`

To solve for x, first factor left side.

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

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

Then, set each factor to zero.

> `cos x - 1=0`

`cos x=1`

`x= 0, 2pi`

> `cos x+1=0`

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`cos^4x-1=0`

To solve for x, first factor left side.

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

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

Then, set each factor to zero.

> `cos x - 1=0`

`cos x=1`

`x= 0, 2pi`

> `cos x+1=0`

`cos x=-1`

`x=pi`

> `cos^2x+1=0`

`cos^2x=-1`

`cos x= sqrt(-1)`

Since the third factor leads to an imaginary value, do not consider it as a solution to the given equation.

Also, there is no indicated interval for the values of x. So, consider the general solution of `cos theta` which is:

`theta = cos^(-1)theta + 2pik ` , where k is any integer.

Replace `cos^-1 theta` with values of x.

`x=0+2pik=2pik`     and        `x= pi + 2pik`

Since k is any integer, we may re-write the two solutions above as one.

`x=pik`

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Therefore, the general solution to the equation `cos^4x-1=0` is `x=pik`.

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