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

Solution steps

2 sin (φ) - 1 = 0

Move

1

to the right side

2 sin (φ) = 1

Divide both sides by

2

sin (φ) = \frac{1}{2}

General solutions for

sin (φ) = \frac{1}{2}

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