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

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

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solvefor θ,2sin(θ)-1=0

solve for

θ, 2 sin (θ) - 1 = 0

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

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) - 1 = 0

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

- 10 sin (x) = - 52

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( b )}{11}

y, \frac{sin ( 7 y )}{7} = \frac{cos ( 2 x )}{2} + \frac{1}{2}

3 cos (\frac{π x}{12}) + 8.4 = 7

sin (x) = \frac{1}{6}, tan (x)

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