What is the general solution for sin^2(θ)-2sin(θ)+1=0 ?

The general solution for sin^2(θ)-2sin(θ)+1=0 is θ= pi/2+2pin

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sin^2(θ)-2sin(θ)+1=0

sin^{2} (θ) - 2 sin (θ) + 1 = 0

θ = \frac{π}{2} + 2 πn

Solution steps

sin^{2} (θ) - 2 sin (θ) + 1 = 0

Solve by substitution

sin (θ) = 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

cos (4 θ) = - \frac{1}{2}

sin (x) = \frac{2 5}{5}

(2 cos (x) - 1) (sin (x) - 1) = 0

cos (3 2^{\circ}) = sin (5 m) - 12

3 = 2 - 2 sin (θ)

What is the general solution for sin^2(θ)-2sin(θ)+1=0 ?
The general solution for sin^2(θ)-2sin(θ)+1=0 is θ= pi/2+2pin