What is the general solution for sin(2θ)+2sin(θ)=0 ?

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

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sin(2θ)+2sin(θ)=0

sin (2 θ) + 2 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

sin (2 θ) + 2 sin (θ) = 0

Rewrite using trig identities

2 sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

2 sin (θ) + 2 cos (θ) sin (θ) : 2 sin (θ) (cos (θ) + 1)

2 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

tan (\frac{x}{2} + π) + 1 = 0

sin (x + \frac{π}{2}) = 1

cos (x π) = - 1

tan (x) = \frac{2}{- 2}

- 6 cos (2 x) = 0

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