Homework Help

Solve: (sinx)^4 + (cosx) ^4 = 1

user profile pic

oldnick | Valedictorian

Posted April 5, 2013 at 12:21 AM via web

dislike 1 like

Solve: (sinx)^4 + (cosx) ^4 = 1

Tagged with math, solve for x, trigonometry

2 Answers | Add Yours

user profile pic

violy | High School Teacher | (Level 3) Assistant Educator

Posted April 5, 2013 at 3:11 AM (Answer #1)

dislike 1 like

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 to 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,..

For sin^2x - 1 = 0, add 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} and x = {3pi/2 + 2kpi}, where k = 0, 1, 2, 3,..

user profile pic

oldnick | Valedictorian

Posted April 6, 2013 at 1:37 PM (Answer #2)

dislike 0 like

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 to 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,..

For sin^2x - 1 = 0, add 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} and x = {3pi/2 + 2kpi}, where k = 0, 1, 2, 3,..


wouldn't be better:

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


on the other side:  sin^2 x + cos^2 x = 1 and,
sin^4 x + cos^4 x = 1

 

thus : 1 = 1 + 2 sin^2 x cos^2 x       

  2 sin^2 x cos ^2 x = 0

4 sin^2x cos^2 x = 0

  (sin2x)^2 = 0    sin 2x = 0      x =  k`pi`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes