What is the general solution for 6cos(θ)=-6(1+cos(θ)) ?

The general solution for 6cos(θ)=-6(1+cos(θ)) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular Trigonometry >

6cos(θ)=-6(1+cos(θ))

Solution

6 cos (θ) = - 6 (1 + cos (θ))

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

6 cos (θ) = - 6 (1 + cos (θ))

Solve by substitution

6 cos (θ) = - 6 (1 + cos (θ))

Let:

cos (θ) = u

6 u = - 6 (1 + u)

6 u = - 6 (1 + u) : u = - \frac{1}{2}

6 u = - 6 (1 + u)

Expand

- 6 (1 + u) : - 6 - 6 u

- 6 (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = - 6, b = 1, c = u

= - 6 \cdot 1 + (- 6) u

Apply minus-plus rules

+ (- a) = - a

= - 6 \cdot 1 - 6 u

Multiply the numbers:

6 \cdot 1 = 6

= - 6 - 6 u

6 u = - 6 - 6 u

Move

6 u

to the left side

6 u = - 6 - 6 u

Add

6 u

to both sides

6 u + 6 u = - 6 - 6 u + 6 u

Simplify

12 u = - 6

12 u = - 6

Divide both sides by

12

12 u = - 6

Divide both sides by

12

\frac{12 u}{12} = \frac{- 6}{12}

Simplify

\frac{12 u}{12} = \frac{- 6}{12}

Simplify

\frac{12 u}{12} : u

\frac{12 u}{12}

Divide the numbers:

\frac{12}{12} = 1

= u

Simplify

\frac{- 6}{12} : - \frac{1}{2}

\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{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{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

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos(θ)=-6(1+cos(θ)) ?
The general solution for 6cos(θ)=-6(1+cos(θ)) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin