What is the general solution for sin^2(x)+cos^4(x)=2 ?

The general solution for sin^2(x)+cos^4(x)=2 is No Solution for x\in\mathbb{R}

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

2

from both sides

sin^{2} (x) + cos^{4} (x) - 2 = 0

Rewrite using trig identities

- 2 + cos^{4} (x) + sin^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 2 + cos^{4} (x) + 1 - cos^{2} (x)

Simplify

- 2 + cos^{4} (x) + 1 - cos^{2} (x) : cos^{4} (x) - cos^{2} (x) - 1

- 2 + cos^{4} (x) + 1 - cos^{2} (x)

Group like terms

= cos^{4} (x) - cos^{2} (x) - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= cos^{4} (x) - cos^{2} (x) - 1

= cos^{4} (x) - cos^{2} (x) - 1

- 1 - cos^{2} (x) + cos^{4} (x) = 0

Solve by substitution

- 1 - cos^{2} (x) + cos^{4} (x) = 0

Let:

cos (x) = u

- 1 - u^{2} + u^{4} = 0

- 1 - u^{2} + u^{4} = 0 : u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

- 1 - u^{2} + u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - v - 1 = 0

Solve

v^{2} - v - 1 = 0 : v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{2} - v - 1 = 0

Solve with the quadratic formula

v^{2} - v - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = - 1

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

(- 1)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

v = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

v = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

\frac{- ( - 1 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{1 + 5}{2} : u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

u^{2} = \frac{1 + 5}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

Solve

u^{2} = \frac{1 - 5}{2} : u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{2} = \frac{1 - 5}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

The solutions are

u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 5}{2}, cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}, cos (x) = - \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2}, cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}, cos (x) = - \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2} :

No Solution

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1 + 5}{2} :

No Solution

cos (x) = - \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1 - 5}{2} : x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn

cos (x) = \frac{1 - 5}{2}

Apply trig inverse properties

cos (x) = \frac{1 - 5}{2}

General solutions for

cos (x) = \frac{1 - 5}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn

x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn

cos (x) = - \frac{1 - 5}{2} : x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

cos (x) = - \frac{1 - 5}{2}

Apply trig inverse properties

cos (x) = - \frac{1 - 5}{2}

General solutions for

cos (x) = - \frac{1 - 5}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

Combine all the solutions

x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn, x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

Since the equation is undefined for:

arccos \frac{1 - 5}{2} + 2 πn, - arccos \frac{1 - 5}{2} + 2 πn, arccos - \frac{1 - 5}{2} + 2 πn, - arccos - \frac{1 - 5}{2} + 2 πn

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)+cos^4(x)=2 ?
The general solution for sin^2(x)+cos^4(x)=2 is No Solution for x\in\mathbb{R}