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

The general solution for 4cos(θ)=4-4cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

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4cos(θ)=4-4cos(θ)

4 cos (θ) = 4 - 4 cos (θ)

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (θ) = 4 - 4 cos (θ)

Solve by substitution

4 cos (θ) = 4 - 4 cos (θ)

Let:

cos (θ) = u

4 u = 4 - 4 u

4 u = 4 - 4 u : u = \frac{1}{2}

4 u = 4 - 4 u

Move

4 u

to the left side

4 u = 4 - 4 u

Add

4 u

to 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

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{π}{3} + 2 πn, θ = \frac{5 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{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin^{2} (x) + 6 sin (x) + 5 = 0

cos (x) = \frac{1}{3}, 0 < x < \frac{π}{2}

3 + 2 sin (θ) = 3 + 2 cos (θ)

\frac{sin ( x )}{cos ( x )} = \frac{1}{3}

5 sin (θ) + 4 = 0

What is the general solution for 4cos(θ)=4-4cos(θ) ?
The general solution for 4cos(θ)=4-4cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin