What is the general solution for cos^2(θ)-cos(θ)-6=0 ?

The general solution for cos^2(θ)-cos(θ)-6=0 is No Solution for θ\in\mathbb{R}

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cos^2(θ)-cos(θ)-6=0

Solution

cos^{2} (θ) - cos (θ) - 6 = 0

Solution

No Solution for θ \in R

Solution steps

cos^{2} (θ) - cos (θ) - 6 = 0

Solve by substitution

cos^{2} (θ) - cos (θ) - 6 = 0

Let:

cos (θ) = u

u^{2} - u - 6 = 0

u^{2} - u - 6 = 0 : u = 3, u = - 2

u^{2} - u - 6 = 0

Solve with the quadratic formula

u^{2} - u - 6 = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = - 6

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

(- 1)^{2} - 4 \cdot 1 \cdot (- 6)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 6

(- 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 6 = 24

4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 5 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 5 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 3, u = - 2

Substitute back

u = cos (θ)

cos (θ) = 3, cos (θ) = - 2

cos (θ) = 3, cos (θ) = - 2

cos (θ) = 3 :

No Solution

cos (θ) = 3

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)-cos(θ)-6=0 ?
The general solution for cos^2(θ)-cos(θ)-6=0 is No Solution for θ\in\mathbb{R}