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

The general solution for 2sin(2θ)-1=0 is θ= pi/(12)+pin,θ=(5pi)/(12)+pin

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2sin(2θ)-1=0

2 sin (2 θ) - 1 = 0

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

cos (\frac{1}{x}) = 0

y, x = sinh (y)

- 3 cos (2 θ) + cos (θ) - 10 = - 6 cos (θ) - 5

7 cot (θ) = 3 csc (θ)

5 sin (x) = 2

What is the general solution for 2sin(2θ)-1=0 ?
The general solution for 2sin(2θ)-1=0 is θ= pi/(12)+pin,θ=(5pi)/(12)+pin