What is the general solution for cos^2(θ)-cos(θ)-1=0 ?

The general solution for cos^2(θ)-cos(θ)-1=0 is θ=2.23703…+2pin,θ=-2.23703…+2pin

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Popular >

cos^2(θ)-cos(θ)-1=0

Solution

cos^{2} (θ) - cos (θ) - 1 = 0

Solution

θ = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

Degrees

θ = 128.17270 \dots^{\circ} + 36 0^{\circ} n, θ = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) - cos (θ) - 1 = 0

Solve by substitution

cos^{2} (θ) - cos (θ) - 1 = 0

Let:

cos (θ) = u

u^{2} - u - 1 = 0

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

u^{2} - u - 1 = 0

Solve with the quadratic formula

u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

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

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

u_{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

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} : \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}

u = \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:

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

Substitute back

u = cos (θ)

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

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

cos (θ) = \frac{1 + 5}{2} :

No Solution

cos (θ) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = \frac{1 - 5}{2}

Apply trig inverse properties

cos (θ) = \frac{1 - 5}{2}

General solutions for

cos (θ) = \frac{1 - 5}{2}

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

θ = arccos (\frac{1 - 5}{2}) + 2 πn, θ = - arccos (\frac{1 - 5}{2}) + 2 πn

θ = arccos (\frac{1 - 5}{2}) + 2 πn, θ = - arccos (\frac{1 - 5}{2}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1 - 5}{2}) + 2 πn, θ = - arccos (\frac{1 - 5}{2}) + 2 πn

Show solutions in decimal form

θ = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)-cos(θ)-1=0 ?
The general solution for cos^2(θ)-cos(θ)-1=0 is θ=2.23703…+2pin,θ=-2.23703…+2pin