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

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

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-2sin(θ)-sin(2θ)=0

- 2 sin (θ) - sin (2 θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

- 2 sin (θ) - sin (2 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

cos (5 x) = - \frac{1}{2}

sin^{2} (x) + 2 sin (x) = - 1

cos (2 x) = 11 sin (x) - 5

x, z = \frac{4}{π} arctan (x y)

sin^{2} (x) cot (x) - cot (x) = 0

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