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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

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

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

Factor

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

sin^{2} (θ) - 2 sin (θ) + 1

Rewrite

sin^{2} (θ) - 2 sin (θ) + 1

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

sin^{2} (θ) - 2 sin (θ) + 1

Rewrite

1

1^{2}

= sin^{2} (θ) - 2 sin (θ) + 1^{2}

Rewrite

2 sin (θ)

2 sin (θ) \cdot 1

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

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = sin (θ), b = 1

= (sin (θ) - 1)^{2}

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

Cancel the common factor:

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 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