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

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

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2sin(φ)-1=0

2 sin (φ) - 1 = 0

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

Degrees

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

Solution steps

2 sin (φ) - 1 = 0

Move

1

to the right side

2 sin (φ) - 1 = 0

Add

1

to both sides

2 sin (φ) - 1 + 1 = 0 + 1

Simplify

2 sin (φ) = 1

2 sin (φ) = 1

Divide both sides by

2

2 sin (φ) = 1

Divide both sides by

2

\frac{2 sin ( φ )}{2} = \frac{1}{2}

Simplify

sin (φ) = \frac{1}{2}

sin (φ) = \frac{1}{2}

General solutions for

sin (φ) = \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}

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

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

sin (x - 2 0^{\circ}) = cos (x)

x, z = sin^{2} (x - 2 π y)

sin (θ) = \frac{3}{3 5}

6 cos (θ) sin (θ) + cos (θ) = 0

cos (x) + cos (5 x) = 0

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