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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)=0,0<= θ<= 2pi

sin (θ) = 0, 0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π

Solution steps

sin (θ) = 0, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 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