# Solve for x if (-sin x)^4+(cos x)^4 =1. Please explain with all steps.

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(-sinx)^4 + (c0sx)^4 = 1

we know that sinx = -sinx

==> (sinx)^4 + (cosx)^4 =1

Complete the square:

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

==> But we know that sin^2 x + cos^2 x = 1

==> 1 - 2sin^2 x *cos^2 x =1

==> -2sin^2 x *cos^2 x = 0

==> sin^2 x *cos^2 x = 0

==> sinx *cosx = 0

**==> sinx = 0 ==> x1= n*pi (n = 0, 1, 2, ...)**

**==> cosx = 0 ==> x2 = (2n+1)pi/2 (n= 0.,1,2...)**

(-sin x)^4+(cos x)^4

=>(sin x)^4+(cos x)^4

To solve this problem we first convert the form of the (sin x)^4+(cos x)^4

Add and subtract 2( sin x cos x)^2

We get:

(sin x)^4+(cos x)^4 + 2(sin x*cos x)^2 - 2(sin x*cos x)^2 =1

=> [(sin x)^2+(cos x)^2]^2 - 2(sin x*cos x)^2 =1

Now we know that (cos x)^2 + (sin x)^2 =1

=> 1 - 2(sin x*cos x)^2 = 1

=> 2(sin x*cos x)^2 = 0

=> (sin x*cos x)^2 = 0

=> (sin x*cos x) = 0

=> 2 sin x cos x =0

=> sin x cos x =0

**=> **sin x = 0 or cos x =0

=> x = n*pi or x = (2n+1)*pi/2

**Therefore the only values of x that solve the equation are**

**x= ****n*pi or x = (2n+1)* pi/2****.**

To solve (-sinx)^4 + (cosx)^4 = 1:

We know (cosx)^2 +(sinx)^2 = 1 , the trigonometric identy.

To solve the equation we put on the right 1= {(sinx)^2+(cosx)^2}^2.

Then (-sinx)^4 +(cosx)^4 = {(sinx)^2+(cosx)^2}^2.

Then (sinx)^4 + (cosx)^4 = (cosx)^4 + 2(sinx)^2*(cosx)^2 +(cosx)^4 , as the coefficient (-1)^4 = 1.

0 = 2(sinx)^2 (cosx)^2. Other terms cancel.

(sinx)^2 = 0. Or( cosx)^2 = 0

sinx = 0. Or cosx = 0

sinx = 0 gives: x = npi, n =0,1,2,...

cosx = 0 gives: x = (2n + or - 1)pi , n = 0, 1,2,.