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

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

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sin(x)-1/2 =0

sin (x) - \frac{1}{2} = 0

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

Degrees

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

Solution steps

sin (x) - \frac{1}{2} = 0

Move

\frac{1}{2}

to the right side

sin (x) - \frac{1}{2} = 0

Add

\frac{1}{2}

to both sides

sin (x) - \frac{1}{2} + \frac{1}{2} = 0 + \frac{1}{2}

Simplify

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

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

General solutions for

sin (x) = \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}

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

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

4 cos^{2} (x) = 3 - 4 cos (x)

sin (x) = 0.32

cos^{4} (x) + sin^{4} (x) = 1

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

sec^{2} (\frac{x}{2}) = 2

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