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 = 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

3 tan^{2} (x) + 4 tan (x) - 4 = 0

csc^{2} (θ) + 2 cot (θ) = 0

sin (x) cos (x) = - 10 cos (x)

tan^{2} (2 x) = 1

8 cos (x) tan (x) = tan (x)

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