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

The general solution for 2sin(θ)-1=0,(0,2pi) is θ= pi/6 ,θ=(5pi)/6

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2sin(θ)-1=0,(0,2pi)

2 sin (θ) - 1 = 0, (0, 2 π)

θ = \frac{π}{6}, θ = \frac{5 π}{6}

Solution steps

2 sin (θ) - 1 = 0, (0, 2 π)

Move

1

to the right side

2 sin (θ) = 1

Divide both sides by

2

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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

Solutions for the range

(0, 2 π)

θ = \frac{π}{6}, θ = \frac{5 π}{6}

a, r = cos (2 a)

cos (2 t) + cos (t) = 0

\frac{π}{6} = sin (3 x)

tan (x) = 3 tan (\frac{1}{2} x)

cos (t) = \frac{- 5}{13}, \frac{1}{2} π < t < π

What is the general solution for 2sin(θ)-1=0,(0,2pi) ?
The general solution for 2sin(θ)-1=0,(0,2pi) is θ= pi/6 ,θ=(5pi)/6