What is the general solution for cos^4(a)=3+4cos^2(a)+cos^4(a) ?

The general solution for cos^4(a)=3+4cos^2(a)+cos^4(a) is No Solution for a\in\mathbb{R}

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Popular >

cos^4(a)=3+4cos^2(a)+cos^4(a)

Solution

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)

Solution

No Solution for a \in R

Solution steps

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)

Solve by substitution

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)

Let:

cos (a) = u

u^{4} = 3 + 4 u^{2} + u^{4}

u^{4} = 3 + 4 u^{2} + u^{4} : u = i \frac{3}{2}, u = - i \frac{3}{2}

u^{4} = 3 + 4 u^{2} + u^{4}

Switch sides

3 + 4 u^{2} + u^{4} = u^{4}

Move

3

to the right side

3 + 4 u^{2} + u^{4} = u^{4}

Subtract

3

from both sides

3 + 4 u^{2} + u^{4} - 3 = u^{4} - 3

Simplify

4 u^{2} + u^{4} = u^{4} - 3

4 u^{2} + u^{4} = u^{4} - 3

Move

u^{4}

to the left side

4 u^{2} + u^{4} = u^{4} - 3

Subtract

u^{4}

from both sides

4 u^{2} + u^{4} - u^{4} = u^{4} - 3 - u^{4}

Simplify

4 u^{2} = - 3

4 u^{2} = - 3

Divide both sides by

4

4 u^{2} = - 3

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{- 3}{4}

Simplify

u^{2} = - \frac{3}{4}

u^{2} = - \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

u = - \frac{3}{4}, u = - - \frac{3}{4}

Simplify

- \frac{3}{4} : i \frac{3}{2}

- \frac{3}{4}

Apply radical rule:

- a = - 1 a

- \frac{3}{4} = - 1 \frac{3}{4}

= - 1 \frac{3}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{3}{4} = \frac{3}{4}

= i \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{3}{2}

Rewrite

i \frac{3}{2}

in standard complex form:

\frac{3}{2} i

i \frac{3}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 i}{2}

= \frac{3}{2} i

Simplify

- - \frac{3}{4} : - i \frac{3}{2}

- - \frac{3}{4}

Simplify

- \frac{3}{4} : i \frac{3}{2}

- \frac{3}{4}

Apply radical rule:

- a = - 1 a

- \frac{3}{4} = - 1 \frac{3}{4}

= - 1 \frac{3}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{3}{4} = \frac{3}{4}

= i \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{3}{2}

= - \frac{3}{2} i

= - i \frac{3}{2}

u = i \frac{3}{2}, u = - i \frac{3}{2}

Substitute back

u = cos (a)

cos (a) = i \frac{3}{2}, cos (a) = - i \frac{3}{2}

cos (a) = i \frac{3}{2}, cos (a) = - i \frac{3}{2}

cos (a) = i \frac{3}{2} :

No Solution

cos (a) = i \frac{3}{2}

No Solution

cos (a) = - i \frac{3}{2} :

No Solution

cos (a) = - i \frac{3}{2}

No Solution

Combine all the solutions

No Solution for a \in R

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1 - cos (x) = 2

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prove sin^2(90-b)+cos^2(b-450)=1 prove

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Frequently Asked Questions (FAQ)

What is the general solution for cos^4(a)=3+4cos^2(a)+cos^4(a) ?
The general solution for cos^4(a)=3+4cos^2(a)+cos^4(a) is No Solution for a\in\mathbb{R}