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

The general solution for cos^2(θ)-2cos(θ)-1=0 is θ=1.99787…+2pin,θ=-1.99787…+2pin

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cos^2(θ)-2cos(θ)-1=0

Solution

cos^{2} (θ) - 2 cos (θ) - 1 = 0

Solution

θ = 1.99787 \dots + 2 πn, θ = - 1.99787 \dots + 2 πn

Degrees

θ = 114.46980 \dots^{\circ} + 36 0^{\circ} n, θ = - 114.46980 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) - 2 cos (θ) - 1 = 0

Solve by substitution

cos^{2} (θ) - 2 cos (θ) - 1 = 0

Let:

cos (θ) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

n ab = n a n b

= 2 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

= \frac{2 ( 1 - 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

u = 1 + 2, u = 1 - 2

Substitute back

u = cos (θ)

cos (θ) = 1 + 2, cos (θ) = 1 - 2

cos (θ) = 1 + 2, cos (θ) = 1 - 2

cos (θ) = 1 + 2 :

No Solution

cos (θ) = 1 + 2

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 1 - 2 : θ = arccos (1 - 2) + 2 πn, θ = - arccos (1 - 2) + 2 πn

cos (θ) = 1 - 2

Apply trig inverse properties

cos (θ) = 1 - 2

General solutions for

cos (θ) = 1 - 2

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

θ = arccos (1 - 2) + 2 πn, θ = - arccos (1 - 2) + 2 πn

θ = arccos (1 - 2) + 2 πn, θ = - arccos (1 - 2) + 2 πn

Combine all the solutions

θ = arccos (1 - 2) + 2 πn, θ = - arccos (1 - 2) + 2 πn

Show solutions in decimal form

θ = 1.99787 \dots + 2 πn, θ = - 1.99787 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)-2cos(θ)-1=0 ?
The general solution for cos^2(θ)-2cos(θ)-1=0 is θ=1.99787…+2pin,θ=-1.99787…+2pin