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

The general solution for 1/2 =sin(2θ) is θ= pi/(12)+pin,θ=(5pi)/(12)+pin

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1/2 =sin(2θ)

\frac{1}{2} = sin (2 θ)

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Solution steps

\frac{1}{2} = sin (2 θ)

Switch sides

sin (2 θ) = \frac{1}{2}

General solutions for

sin (2 θ) = \frac{1}{2}

2 θ = \frac{π}{6} + 2 πn, 2 θ = \frac{5 π}{6} + 2 πn

Solve

2 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{12} + πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

1 - 2 sin (θ) = - 1

sin (2 x) + sin (x) = sin (3 x)

sin (2 x) = 23 sin^{2} (x)

cosh (x) = 5

cosh (x) = 7

What is the general solution for 1/2 =sin(2θ) ?
The general solution for 1/2 =sin(2θ) is θ= pi/(12)+pin,θ=(5pi)/(12)+pin