# THE GENERAL SOLUTION OF THE EQUATION SIN^4X + COS^4X= SINX

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"Solve: (sinx)^4 + (cosx) ^4 = 1"*buoidienxanhha.com Editorial*, 5 Apr. 2013, https://www.buoidienxanhha.com/homework-help/what-about-senx-4-cosx-4-1-solutions-425700.Accessed 24 Dec. 2022.

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We can rewrite the equation as:

sin^4x + cos^4x = 1

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

So, we will 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 by 1 - sin^2x on our equation.

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

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

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

Combine like 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 factor khổng lồ zero.

sin^2x = 0

Take the square root of both sides.

sinx = 0

So,** x = 0, pi, 2pi in interval <0, 2pi>. For the general solutions: **

**x = 0 + 2kpi, pi + 2kpi, 2pi + 2kpi. Where k =0, 1,2,3,..Xem thêm: 6 Reasons Why Knowing English Is Important In Today'S World**

For sin^2x - 1 = 0, địa chỉ 1 on both sides.

sin^2x = 1

Take the square root of both sides.

sinx = +/- 1.

So, **x = pi/2, 3pi/2 in interval <0, 2pi>. For the general solutions:**

**x = pi/2 + 2kpi & x = 3pi/2 + 2kpi, where k = 0, 1, 2, 3,..**

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