What is the general solution for 8cos(2θ)=-24cos(θ)-16 ?

The general solution for 8cos(2θ)=-24cos(θ)-16 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

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8cos(2θ)=-24cos(θ)-16

Solution

8 cos (2 θ) = - 24 cos (θ) - 16

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

8 cos (2 θ) = - 24 cos (θ) - 16

Subtract

- 24 cos (θ) - 16

from both sides

8 cos (2 θ) + 24 cos (θ) + 16 = 0

Rewrite using trig identities

16 + 24 cos (θ) + 8 cos (2 θ)

Use the Double Angle identity:

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

= 16 + 24 cos (θ) + 8 (2 cos^{2} (θ) - 1)

Simplify

16 + 24 cos (θ) + 8 (2 cos^{2} (θ) - 1) : 16 cos^{2} (θ) + 24 cos (θ) + 8

16 + 24 cos (θ) + 8 (2 cos^{2} (θ) - 1)

Expand

8 (2 cos^{2} (θ) - 1) : 16 cos^{2} (θ) - 8

8 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 2 cos^{2} (θ), c = 1

= 8 \cdot 2 cos^{2} (θ) - 8 \cdot 1

Simplify

8 \cdot 2 cos^{2} (θ) - 8 \cdot 1 : 16 cos^{2} (θ) - 8

8 \cdot 2 cos^{2} (θ) - 8 \cdot 1

Multiply the numbers:

8 \cdot 2 = 16

= 16 cos^{2} (θ) - 8 \cdot 1

Multiply the numbers:

8 \cdot 1 = 8

= 16 cos^{2} (θ) - 8

= 16 cos^{2} (θ) - 8

= 16 + 24 cos (θ) + 16 cos^{2} (θ) - 8

Simplify

16 + 24 cos (θ) + 16 cos^{2} (θ) - 8 : 16 cos^{2} (θ) + 24 cos (θ) + 8

16 + 24 cos (θ) + 16 cos^{2} (θ) - 8

Group like terms

= 24 cos (θ) + 16 cos^{2} (θ) + 16 - 8

Add/Subtract the numbers:

16 - 8 = 8

= 16 cos^{2} (θ) + 24 cos (θ) + 8

= 16 cos^{2} (θ) + 24 cos (θ) + 8

= 16 cos^{2} (θ) + 24 cos (θ) + 8

8 + 16 cos^{2} (θ) + 24 cos (θ) = 0

Solve by substitution

8 + 16 cos^{2} (θ) + 24 cos (θ) = 0

Let:

cos (θ) = u

8 + 16 u^{2} + 24 u = 0

8 + 16 u^{2} + 24 u = 0 : u = - \frac{1}{2}, u = - 1

8 + 16 u^{2} + 24 u = 0

Write in the standard form

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

16 u^{2} + 24 u + 8 = 0

Solve with the quadratic formula

16 u^{2} + 24 u + 8 = 0

Quadratic Equation Formula:

For

a = 16, b = 24, c = 8

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

2 4^{2} - 4 \cdot 16 \cdot 8 = 8

2 4^{2} - 4 \cdot 16 \cdot 8

Multiply the numbers:

4 \cdot 16 \cdot 8 = 512

= 2 4^{2} - 512

2 4^{2} = 576

= 576 - 512

Subtract the numbers:

576 - 512 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 24 \pm 8}{2 \cdot 16}

Separate the solutions

u_{1} = \frac{- 24 + 8}{2 \cdot 16}, u_{2} = \frac{- 24 - 8}{2 \cdot 16}

u = \frac{- 24 + 8}{2 \cdot 16} : - \frac{1}{2}

\frac{- 24 + 8}{2 \cdot 16}

Add/Subtract the numbers:

- 24 + 8 = - 16

= \frac{- 16}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 16}{32}

Apply the fraction rule:

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

= - \frac{16}{32}

Cancel the common factor:

16

= - \frac{1}{2}

u = \frac{- 24 - 8}{2 \cdot 16} : - 1

\frac{- 24 - 8}{2 \cdot 16}

Subtract the numbers:

- 24 - 8 = - 32

= \frac{- 32}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 32}{32}

Apply the fraction rule:

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

= - \frac{32}{32}

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 8cos(2θ)=-24cos(θ)-16 ?
The general solution for 8cos(2θ)=-24cos(θ)-16 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin