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

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

`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`.

salonigaba | Student, Grade 11 | (Level 1) Salutatorian

Posted on

cos^4x-1=0

cos^2x-1*cos^2x+1=0

cosx-1*cosx+1*cos^2x+1=0

cosx-1=0

cosx=1

x=0,2pie

cosx=-1

x=pie

cos^2+1=0

cos^2=-1

cos=sqrrt-1

co it will not consider

thats's why: cos^4x-1=0

x=pie*K