What is the general solution for 2cos(θ)-1=0 ?

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

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2cos(θ)-1=0

2 cos (θ) - 1 = 0

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

2 cos (θ) - 1 = 0

Move

1

to the right side

2 cos (θ) - 1 = 0

Add

1

to both sides

2 cos (θ) - 1 + 1 = 0 + 1

Simplify

2 cos (θ) = 1

2 cos (θ) = 1

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

sin (x) = 0.4

cos (\frac{3 π}{8})

tan (θ) + 1 = 0

arctan (\frac{π}{2})

sin (- \frac{3 π}{2})

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