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

The general solution for sin(b)=(sqrt(2))/2 is b= pi/4+2pin,b=(3pi)/4+2pin

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sin(b)=(sqrt(2))/2

sin (b) = \frac{2}{2}

b = \frac{π}{4} + 2 πn, b = \frac{3 π}{4} + 2 πn

Degrees

b = 4 5^{\circ} + 36 0^{\circ} n, b = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (b) = \frac{2}{2}

General solutions for

sin (b) = \frac{2}{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}

b = \frac{π}{4} + 2 πn, b = \frac{3 π}{4} + 2 πn

b = \frac{π}{4} + 2 πn, b = \frac{3 π}{4} + 2 πn

h = 7 cos (\frac{π}{3} t), t

3 + 3 sin (θ) = 3 + 3 cos (θ)

- 3 = tan (θ)

cot (θ) = 13

arctan (0.2 x) + arctan (0.0625 x) = 5 4^{\circ}

What is the general solution for sin(b)=(sqrt(2))/2 ?
The general solution for sin(b)=(sqrt(2))/2 is b= pi/4+2pin,b=(3pi)/4+2pin