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

The general solution for sin^2(2θ)=1 is θ= pi/4+pin,θ=(3pi)/4+pin

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sin^2(2θ)=1

sin^{2} (2 θ) = 1

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Solution steps

sin^{2} (2 θ) = 1

Solve by substitution

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

sin (2 θ) = - 1 : θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

sin (x) = \frac{6}{7}

x, sin (x) = - \frac{1}{2}

sin (\frac{θ}{2}) = \frac{1}{2}

4 cos (θ) - 22 = 0

4 cos (4 x) = 0

What is the general solution for sin^2(2θ)=1 ?
The general solution for sin^2(2θ)=1 is θ= pi/4+pin,θ=(3pi)/4+pin