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

The general solution for sin(θ)=0,0<= θ<= 2pi is θ=0,θ=pi,θ=2pi

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sin(θ)=0,0<= θ<= 2pi

sin (θ) = 0, 0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 36 0^{\circ}

Solution steps

sin (θ) = 0, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 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, θ = π + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π

sin (θ) = - 0.71

2 cos (4 x) = 0

tan (2 θ) = \frac{- 1}{1}

cos^{2} (\frac{x}{2}) = \frac{1}{2}

5^{5 - 3 c s c (θ)} = 1

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