What is the general solution for -4cos(θ)=4(1+cos(θ)) ?

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

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-4cos(θ)=4(1+cos(θ))

Solution

- 4 cos (θ) = 4 (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

- 4 cos (θ) = 4 (1 + cos (θ))

Solve by substitution

- 4 cos (θ) = 4 (1 + cos (θ))

Let:

cos (θ) = u

- 4 u = 4 (1 + u)

- 4 u = 4 (1 + u) : u = - \frac{1}{2}

- 4 u = 4 (1 + u)

Expand

4 (1 + u) : 4 + 4 u

4 (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = 1, c = u

= 4 \cdot 1 + 4 u

Multiply the numbers:

4 \cdot 1 = 4

= 4 + 4 u

- 4 u = 4 + 4 u

Move

4 u

to the left side

- 4 u = 4 + 4 u

Subtract

4 u

from both sides

- 4 u - 4 u = 4 + 4 u - 4 u

Simplify

- 8 u = 4

- 8 u = 4

Divide both sides by

- 8

- 8 u = 4

Divide both sides by

- 8

\frac{- 8 u}{- 8} = \frac{4}{- 8}

Simplify

\frac{- 8 u}{- 8} = \frac{4}{- 8}

Simplify

\frac{- 8 u}{- 8} : u

\frac{- 8 u}{- 8}

Apply the fraction rule:

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

= \frac{8 u}{8}

Divide the numbers:

\frac{8}{8} = 1

= u

Simplify

\frac{4}{- 8} : - \frac{1}{2}

\frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \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

Popular Examples

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sin (2 a) = 1

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tan (2 x) - \frac{1}{tan ( x )} = 0

sin(x)+cos(x)+1=0

sin (x) + cos (x) + 1 = 0

tan(x)=15

tan (x) = 15

4sin(x)=cos(x)+2

4 sin (x) = cos (x) + 2

Frequently Asked Questions (FAQ)

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