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

The general solution for sqrt(2)sin(2θ)=1 is θ= pi/8+pin,θ=(3pi)/8+pin

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sqrt(2)sin(2θ)=1

2 sin (2 θ) = 1

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

Solution steps

2 sin (2 θ) = 1

Divide both sides by

2

sin (2 θ) = \frac{2}{2}

General solutions for

sin (2 θ) = \frac{2}{2}

2 θ = \frac{π}{4} + 2 πn, 2 θ = \frac{3 π}{4} + 2 πn

Solve

2 θ = \frac{π}{4} + 2 πn : θ = \frac{π}{8} + πn

Solve

2 θ = \frac{3 π}{4} + 2 πn : θ = \frac{3 π}{8} + πn

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

sin (x) a = \frac{π}{2}

tan^{2} (3 θ) = 1

tan^{2} (2 θ) - 3 = 0

cos (2 x) = 5 sin^{2} (x) - cos^{2} (x)

What is the general solution for sqrt(2)sin(2θ)=1 ?
The general solution for sqrt(2)sin(2θ)=1 is θ= pi/8+pin,θ=(3pi)/8+pin