What is the general solution for sin(y)=(sqrt(3))/2 ?

The general solution for sin(y)=(sqrt(3))/2 is y= pi/3+2pin,y=(2pi)/3+2pin

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sin(y)=(sqrt(3))/2

sin (y) = \frac{3}{2}

y = \frac{π}{3} + 2 πn, y = \frac{2 π}{3} + 2 πn

Degrees

y = 6 0^{\circ} + 36 0^{\circ} n, y = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (y) = \frac{3}{2}

General solutions for

sin (y) = \frac{3}{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}

y = \frac{π}{3} + 2 πn, y = \frac{2 π}{3} + 2 πn

y = \frac{π}{3} + 2 πn, y = \frac{2 π}{3} + 2 πn

9 sin^{2} (x) + 3 cos (x) - 7 = 0

A, sin (B + \frac{C}{2}) = cos (\frac{A}{2})

33 tan (u) = 3

3 sec (2 x + 3) = 4

t, f = 3 cos (ptt h e t a) - 2

What is the general solution for sin(y)=(sqrt(3))/2 ?
The general solution for sin(y)=(sqrt(3))/2 is y= pi/3+2pin,y=(2pi)/3+2pin