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

The general solution for cos^2(θ)-cos(θ)-2=0 is θ=pi+2pin

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cos^2(θ)-cos(θ)-2=0

Solution

cos^{2} (θ) - cos (θ) - 2 = 0

Solution

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) - cos (θ) - 2 = 0

Solve by substitution

cos^{2} (θ) - cos (θ) - 2 = 0

Let:

cos (θ) = u

u^{2} - u - 2 = 0

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

u^{2} - u - 2 = 0

Solve with the quadratic formula

u^{2} - u - 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 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 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 3 = - 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:

u = 2, u = - 1

Substitute back

u = cos (θ)

cos (θ) = 2, cos (θ) = - 1

cos (θ) = 2, cos (θ) = - 1

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = π + 2 πn

Popular Examples

2sin(3x)=-1

2 sin (3 x) = - 1

8sin^2(θ)+2sin(θ)-1=0

8 sin^{2} (θ) + 2 sin (θ) - 1 = 0

tan(θ)=6

tan (θ) = 6

sin(4x)= 1/2

sin (4 x) = \frac{1}{2}

cos(x)-sin(x)=0,0<= x<2pi

cos (x) - sin (x) = 0, 0 \leq x < 2 π

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)-cos(θ)-2=0 ?
The general solution for cos^2(θ)-cos(θ)-2=0 is θ=pi+2pin