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

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

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4cos(θ)=2

4 cos (θ) = 2

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) = 2

Divide both sides by

4

4 cos (θ) = 2

Divide both sides by

4

\frac{4 cos ( θ )}{4} = \frac{2}{4}

Simplify

cos (θ) = \frac{1}{2}

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

20 cos^{6} (x) - 57 cos^{4} (x) + 27 cos^{2} (x) = 0

2 sin^{2} (w) - 3 sin (w) + 1 = 0

2 + sin (x) = 5 sin (x)

2 cos (θ) + 5 = 0

sin (x) cos (x) = cos (x)

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