What is the general solution for (sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1 ?

The general solution for (sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1 is θ=2pin,θ=pi+2pin

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(sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

θ = 2 πn, θ = π + 2 πn

Solution steps

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

Simplify

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} : sin (θ) - 1

sin (θ) - 1 = - 1

Add

1

to both sides

sin (θ) - 1 + 1 = - 1 + 1

Simplify

sin (θ) = 0

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn, θ = π + 2 πn

w, 100 cos (10 w) = 50

cos (θ) = \frac{8}{25}

sin (2 x) = π

sin (x + 5 π) = 0

cos (2 x) + 9 cos (x) + 5 = 0

What is the general solution for (sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1 ?
The general solution for (sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1 is θ=2pin,θ=pi+2pin