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

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

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-2sin(θ)+2sin^2(θ)=0

- 2 sin (θ) + 2 sin^{2} (θ) = 0

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

Solution steps

- 2 sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = 0

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

sin (2 π x) cos (π x) + cos (2 π x) sin (π x) = - 1

0 = - (sin (6.9256 - x) + (sin (- x)))

- 4 π^{2} sin (2 π x) = 0

- 9 sin (y) = 0

6 + 4 cos (2 x) = 2 - 4 cos (x)

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