What is the general solution for cos^4(x)-3cos^2(x)-4=0 ?

The general solution for cos^4(x)-3cos^2(x)-4=0 is No Solution for x\in\mathbb{R}

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cos^4(x)-3cos^2(x)-4=0

Solution

cos^{4} (x) - 3 cos^{2} (x) - 4 = 0

Solution

No Solution for x \in R

Solution steps

cos^{4} (x) - 3 cos^{2} (x) - 4 = 0

Solve by substitution

cos^{4} (x) - 3 cos^{2} (x) - 4 = 0

Let:

cos (x) = u

u^{4} - 3 u^{2} - 4 = 0

u^{4} - 3 u^{2} - 4 = 0 : u = 2, u = - 2, u = i, u = - i

u^{4} - 3 u^{2} - 4 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - 3 v - 4 = 0

Solve

v^{2} - 3 v - 4 = 0 : v = 4, v = - 1

v^{2} - 3 v - 4 = 0

Solve with the quadratic formula

v^{2} - 3 v - 4 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = - 4

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

(- 3)^{2} - 4 \cdot 1 \cdot (- 4)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 1 \cdot 4

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 4

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

v_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

v = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

\frac{- ( - 3 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Add the numbers:

3 + 5 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

v = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Subtract the numbers:

3 - 5 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

v = 4, v = - 1

v = 4, v = - 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

For

x^{2} = f (a)

the solutions are

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

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

Solve

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 2, u = - 2, u = i, u = - i

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = - 2, cos (x) = i, cos (x) = - i

cos (x) = 2, cos (x) = - 2, cos (x) = i, cos (x) = - i

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos^4(x)-3cos^2(x)-4=0 ?
The general solution for cos^4(x)-3cos^2(x)-4=0 is No Solution for x\in\mathbb{R}