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

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### 2 Answers

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,..**

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`