What is the general solution for 2cos(θ)=1,0<= θ<= 2pi ?

The general solution for 2cos(θ)=1,0<= θ<= 2pi is θ= pi/3 ,θ=(5pi)/3

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2cos(θ)=1,0<= θ<= 2pi

2 cos (θ) = 1, 0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

2 cos (θ) = 1, 0 \leq θ \leq 2 π

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

arctan (x + 1) + arctan (x - 1) = arctan (\frac{8}{31})

4 cos (2 θ) + 19 = - 22 cos (θ) + 6

8 cos^{2} (x) + 16 cos (x) + 8 = 0

cos (x) = - 0.35

cos (x) = - 0.25

What is the general solution for 2cos(θ)=1,0<= θ<= 2pi ?
The general solution for 2cos(θ)=1,0<= θ<= 2pi is θ= pi/3 ,θ=(5pi)/3