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,..

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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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