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

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

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-sin(θ)-sin(2θ)=0

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

cot (θ) = 1.5

- 1 = 2 sin (x)

- 6 sin^{2} (θ) - 2 sin (θ) - 3 = 3 sin (θ) - 2

arctan (x) + arctan (\frac{1}{x}) = \frac{π}{2}

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