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

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

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sin^2(2θ)=0

sin^{2} (2 θ) = 0

θ = πn, θ = \frac{π}{2} + πn

Solution steps

sin^{2} (2 θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (2 θ) = 0

General solutions for

sin (2 θ) = 0

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

y, x = 2 sec (y)

tan (π + c) = 1

8 sin^{2} (x) = 8 cos^{2} (x)

sin (x) = \frac{6}{8}

tan (x) = 2 + tan (3 x)

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