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}

Degrees

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Solution steps

2 sin (θ) - 1 = 0, (0, 2 π)

Move

1

to the right side

2 sin (θ) - 1 = 0

Add

1

to both sides

2 sin (θ) - 1 + 1 = 0 + 1

Simplify

2 sin (θ) = 1

2 sin (θ) = 1

Divide both sides by

2

2 sin (θ) = 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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