What is the general solution for 2cos^2(θ)-cos(θ)-4=-6cos(θ)-6 ?

The general solution for 2cos^2(θ)-cos(θ)-4=-6cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos^2(θ)-cos(θ)-4=-6cos(θ)-6

Solution

2 cos^{2} (θ) - cos (θ) - 4 = - 6 cos (θ) - 6

Solution

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (θ) - cos (θ) - 4 = - 6 cos (θ) - 6

Solve by substitution

2 cos^{2} (θ) - cos (θ) - 4 = - 6 cos (θ) - 6

Let:

cos (θ) = u

2 u^{2} - u - 4 = - 6 u - 6

2 u^{2} - u - 4 = - 6 u - 6 : u = - \frac{1}{2}, u = - 2

2 u^{2} - u - 4 = - 6 u - 6

Move

6

to the left side

2 u^{2} - u - 4 = - 6 u - 6

Add

6

to both sides

2 u^{2} - u - 4 + 6 = - 6 u - 6 + 6

Simplify

2 u^{2} - u + 2 = - 6 u

2 u^{2} - u + 2 = - 6 u

Move

6 u

to the left side

2 u^{2} - u + 2 = - 6 u

Add

6 u

to both sides

2 u^{2} - u + 2 + 6 u = - 6 u + 6 u

Simplify

2 u^{2} + 5 u + 2 = 0

2 u^{2} + 5 u + 2 = 0

Solve with the quadratic formula

2 u^{2} + 5 u + 2 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 2 = 3

5^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 3}{2 \cdot 2}, u_{2} = \frac{- 5 - 3}{2 \cdot 2}

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

\frac{- 5 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 5 - 3}{2 \cdot 2} : - 2

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

Subtract the numbers:

- 5 - 3 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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 = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-cos(θ)-4=-6cos(θ)-6 ?
The general solution for 2cos^2(θ)-cos(θ)-4=-6cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin