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

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

tan (β) = \frac{1}{2}

cos (B) = \frac{4}{5}

2 sin^{2} (x) - 3 cos (2 x) = 3

cos (π + x) + sin (x - \frac{π}{2}) = 1

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