What is the general solution for sin(θ)cos(θ)+sin(θ)=0 ?

The general solution for sin(θ)cos(θ)+sin(θ)=0 is θ=2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)cos(θ)+sin(θ)=0

sin (θ) cos (θ) + sin (θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

sin (θ) cos (θ) + sin (θ) = 0

Factor

sin (θ) cos (θ) + sin (θ) : sin (θ) (cos (θ) + 1)

sin (θ) (cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos (θ) + 1 = 0 : θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

6 cos^{2} (x) - cos (x) - 1 = 0

45.6 = arcsin (\frac{n ^{1}}{n ^{2}}), n^{1}

tan (x) = 0.839

sin (\frac{3 ( π - x )}{2}) = 0

sin (x) (2 - sin (x)) = cos^{2} (x)

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