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

The general solution for sin(2θ)+sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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sin(2θ)+sin(θ)=0

sin (2 θ) + sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Solution steps

sin (2 θ) + sin (θ) = 0

Rewrite using trig identities

sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

sin (θ) (2 cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

2 cos (θ) + 1 = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = \frac{3}{2}

tan (- π)

sin (x) = - 0.7

cot (3 0^{\circ})

\frac{cos ( 2 x )}{( cos ( x ) + sin ( x ) )} = cos (x) - sin (x)

What is the general solution for sin(2θ)+sin(θ)=0 ?
The general solution for sin(2θ)+sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin