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

The general solution for 1-sin^2(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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1-sin^2(θ)=0

1 - sin^{2} (θ) = 0

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

1 - sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

x, z = y sin (x y)

2 sin^{2} (x) + csc^{2} (x) = 3

t, tan (t) = 2

sin (α) = \frac{8}{17}

cos (x) = - sin (\frac{π}{3})

What is the general solution for 1-sin^2(θ)=0 ?
The general solution for 1-sin^2(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin