# What x (senx) ^4 - (cosx)^4 =1proved to be like out of solutions?

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Take note that (sinx)^4 = sin^4x, and (cos)^4 = cos^4x.

So, we will have: sin^4x - cos^4x = 1

We can factor the left side:

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

Note that sin^2x + cos^2x = 1.

So, the equation will now becomes: sin^2x - cos^2x = 1.

Subtract both sides by 1.

sin^2x - cos^2x - 1 = 0

We can rewrite that as: -1 + sin^2x - cos^2x = 0.

Factor out -1 from -1 + sin^2x.

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

-1cos^2x - cos^2x = 0 (since, sin^2x + cos^2x = 1).

Combining like terms.

-2cos^2x = 0

Divide both sides by -2

cos^2x = 0

Take the square root of both sides.

cosx = 0

So,** x ={pi/2, 3pi/2}**.

`(sinx)^4-(cosx)^4=1`

`((sinx)^2)^2-((cosx)^2)^2=1`

`` Factoring by `A^2-B^2=(A-B)(A+B)`

`((sinx)^2-(cosx)^2)((sinx)^2+(cosx)^2)=1`

`sin^2x-cos^2x-1=0`

since `sin^2x+cos^2x=1`

`sin^2x-1-cos^2x=0`

`-cos^2x-cos^2x=0`

`-2cos^2=0`

`cosx=0`

`x=2npi+-pi/2`

``where n is an integer.

OR

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

`-cos(2x)=1`

`cos(2x)=-1`

`cos(2x)=cos(pi)`

`2x=2npi+-pi`

`x=npi+-pi/2`

n is an integer

Take note that (sinx)^4 = sin^4x, and (cos)^4 = cos^4x.

So, we will have: sin^4x - cos^4x = 1

We can factor the left side:

(sin^2x - cos^2x)(sin^2x + cos^2x) = 1Note that sin^2x + cos^2x = 1.

So, the equation will now becomes: sin^2x - cos^2x = 1.Subtract both sides by 1.

sin^2x - cos^2x - 1 = 0

We can rewrite that as: -1 + sin^2x - cos^2x = 0.

Factor out -1 from -1 + sin^2x.-1(1 - sin^2x) - cos^2x = 0

-1cos^2x - cos^2x = 0 (since, sin^2x + cos^2x = 1).Combining like terms.

-2cos^2x = 0Divide both sides by -2

cos^2x = 0

Take the square root of both sides.

cosx = 0So,

x ={pi/2, 3pi/2}.

Would not it be better from: sin^2x - cos^2x =1

(sin x + cos x)(sin x - cos x) =1

raising both members to the square:

(1 + 2sin x cos x)( 1 - 2 sin x cos x) =1

(1 + sin 2x)( 1 - sin 2x) =1

1 - (sin 2x) ^2 =1 sin 2x = 0

x = k`pi` /2 , ( k =1,2.....) ?