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

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

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2sin(a)-1=0

2 sin (a) - 1 = 0

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

Degrees

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

Solution steps

2 sin (a) - 1 = 0

Move

1

to the right side

2 sin (a) - 1 = 0

Add

1

to both sides

2 sin (a) - 1 + 1 = 0 + 1

Simplify

2 sin (a) = 1

2 sin (a) = 1

Divide both sides by

2

2 sin (a) = 1

Divide both sides by

2

\frac{2 sin ( a )}{2} = \frac{1}{2}

Simplify

sin (a) = \frac{1}{2}

sin (a) = \frac{1}{2}

General solutions for

sin (a) = \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}

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

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

cos (3 x) = cos (2 x) cos (x)

3 cos^{2} (x) + sin^{2} (x) + 5 sin (x) = 0

3 cos (θ) = 2 - sin (θ)

2 cos^{2} (θ) sin (θ) + cos^{2} (θ) = 0

cos^{2} (x) tan (x) = tan (x)

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