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

The general solution for sqrt(2)sin(2θ)-1=0 is θ= pi/8+pin,θ=(3pi)/8+pin

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sqrt(2)sin(2θ)-1=0

2 sin (2 θ) - 1 = 0

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

Solution steps

2 sin (2 θ) - 1 = 0

Move

1

to the right side

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

3 sin (θ) = 2 - sin (θ)

tan^{2} (3 t + \frac{π}{2}) = \frac{1}{3}

0 = - 2 sin (2 x)

2 sin^{2} (θ) - 7 sin (θ) + 5 = 0

x, 4 cos^{2} (x) - 3 = 0

What is the general solution for sqrt(2)sin(2θ)-1=0 ?
The general solution for sqrt(2)sin(2θ)-1=0 is θ= pi/8+pin,θ=(3pi)/8+pin