What is the general solution for 3cos(2θ)=-9cos(θ)-6 ?

The general solution for 3cos(2θ)=-9cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

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3cos(2θ)=-9cos(θ)-6

Solution

3 cos (2 θ) = - 9 cos (θ) - 6

Solution

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (2 θ) = - 9 cos (θ) - 6

Subtract

- 9 cos (θ) - 6

from both sides

3 cos (2 θ) + 9 cos (θ) + 6 = 0

Rewrite using trig identities

6 + 3 cos (2 θ) + 9 cos (θ)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= 6 + 3 (2 cos^{2} (θ) - 1) + 9 cos (θ)

Simplify

6 + 3 (2 cos^{2} (θ) - 1) + 9 cos (θ) : 6 cos^{2} (θ) + 9 cos (θ) + 3

6 + 3 (2 cos^{2} (θ) - 1) + 9 cos (θ)

Expand

3 (2 cos^{2} (θ) - 1) : 6 cos^{2} (θ) - 3

3 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (θ), c = 1

= 3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Simplify

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1 : 6 cos^{2} (θ) - 3

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 2 = 6

= 6 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 6 cos^{2} (θ) - 3

= 6 cos^{2} (θ) - 3

= 6 + 6 cos^{2} (θ) - 3 + 9 cos (θ)

Simplify

6 + 6 cos^{2} (θ) - 3 + 9 cos (θ) : 6 cos^{2} (θ) + 9 cos (θ) + 3

6 + 6 cos^{2} (θ) - 3 + 9 cos (θ)

Group like terms

= 6 cos^{2} (θ) + 9 cos (θ) + 6 - 3

Add/Subtract the numbers:

6 - 3 = 3

= 6 cos^{2} (θ) + 9 cos (θ) + 3

= 6 cos^{2} (θ) + 9 cos (θ) + 3

= 6 cos^{2} (θ) + 9 cos (θ) + 3

3 + 6 cos^{2} (θ) + 9 cos (θ) = 0

Solve by substitution

3 + 6 cos^{2} (θ) + 9 cos (θ) = 0

Let:

cos (θ) = u

3 + 6 u^{2} + 9 u = 0

3 + 6 u^{2} + 9 u = 0 : u = - \frac{1}{2}, u = - 1

3 + 6 u^{2} + 9 u = 0

Write in the standard form

a x^{2} + b x + c = 0

6 u^{2} + 9 u + 3 = 0

Solve with the quadratic formula

6 u^{2} + 9 u + 3 = 0

Quadratic Equation Formula:

For

a = 6, b = 9, c = 3

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

9^{2} - 4 \cdot 6 \cdot 3 = 3

9^{2} - 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

Subtract the numbers:

81 - 72 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 9 \pm 3}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 9 + 3}{2 \cdot 6}, u_{2} = \frac{- 9 - 3}{2 \cdot 6}

u = \frac{- 9 + 3}{2 \cdot 6} : - \frac{1}{2}

\frac{- 9 + 3}{2 \cdot 6}

Add/Subtract the numbers:

- 9 + 3 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

u = \frac{- 9 - 3}{2 \cdot 6} : - 1

\frac{- 9 - 3}{2 \cdot 6}

Subtract the numbers:

- 9 - 3 = - 12

= \frac{- 12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 12}{12}

Apply the fraction rule:

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

= - \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

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

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(2θ)=-9cos(θ)-6 ?
The general solution for 3cos(2θ)=-9cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin