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

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

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cos(θ)=(1/2)

cos (θ) = (\frac{1}{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

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^{2} (x) - cos^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π

cos (θ) = - cos (2 θ)

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

\frac{sin ( 2 x )}{sin ( x )} = 1

θ, ma = t sin (θ) - m g

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