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

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

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sqrt(2)=2sin(2θ)

2 = 2 sin (2 θ)

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

Solution steps

2 = 2 sin (2 θ)

Switch sides

2 sin (2 θ) = 2

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

4 sin^{2} (x) + 4 cos (x) - 5 = 0, (0, 2 π)

a, E = 25 (sin (a) - cos (a))

391 = 78 \cdot 21.5 \cdot cos (x)

5 tan (θ) - 9 = - 6 tan (θ) + 8

81 (cos^{2} (x)) = 6

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