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You are watching: Sin^4x+cos^4x We have the right to rewrite the equation as:

sin^4x + cos^4x = 1

Take note that cos^4x = (cos^2x)^2.

So, we will certainly have:

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

Using the identity sin^2x + cos^2x = 1, we will have:

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

replace the cos^2x through 1 - sin^2x on...

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We have the right to rewrite the equation as:

sin^4x + cos^4x = 1

Take note that cos^4x = (cos^2x)^2.

So, we will certainly have:

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

Using the identity sin^2x + cos^2x = 1, we will certainly have:

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

Replace the cos^2x through 1 - sin^2x on ours equation.

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

Using silver paper on the (1 - sin^2x)^2.

sin^4x + 1 - 2sin^2x + sin^4x = 1

Combine choose terms.

2sin^4x - 2sinx + 1 = 1

Subtract both sides by 1.

2sin^4x - 2sin^2x = 0

Divide both sides by 2.

sin^4x - sin^2x = 0

Factor the left side.

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

Equate each variable to zero.

sin^2x = 0

Take the square root of both sides.

sinx = 0

So, x = 0, pi, 2pi in term <0, 2pi>. Because that the general solutions:

x = 0 + 2kpi, pi + 2kpi, 2pi + 2kpi. Wherein k =0, 1,2,3,..

For sin^2x - 1 = 0, include 1 top top both sides.

sin^2x = 1

Take the square root of both sides.

sinx = +/- 1.

So, x = pi/2, 3pi/2 in interval <0, 2pi>. Because that the basic solutions:

x = pi/2 + 2kpi and also x = 3pi/2 + 2kpi, where k = 0, 1, 2, 3,..

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