What is the general solution for 2+2cos(θ)=4 ?

The general solution for 2+2cos(θ)=4 is θ=2pin

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2+2cos(θ)=4

2 + 2 cos (θ) = 4

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 + 2 cos (θ) = 4

Move

2

to the right side

2 + 2 cos (θ) = 4

Subtract

2

from both sides

2 + 2 cos (θ) - 2 = 4 - 2

Simplify

2 cos (θ) = 2

2 cos (θ) = 2

Divide both sides by

2

2 cos (θ) = 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{2}{2}

Simplify

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sin^{2} (x) - 2 sin (x) = 3

sin (x + \frac{π}{3}) = \frac{1}{2}

\frac{sin ( 4 7 ^{\circ} )}{12} = \frac{sin ( x )}{11}

cos (2 x) + 3 cos (x) = - 1

2 cos (α) = 3

What is the general solution for 2+2cos(θ)=4 ?
The general solution for 2+2cos(θ)=4 is θ=2pin