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

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

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(sqrt(2))/2 =sin(x)

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

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

Degrees

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

Solution steps

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

Switch sides

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

General solutions for

sin (x) = \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}

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

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

tan (θ) = \frac{12}{6}

tan (θ) = \frac{12}{7}

2 sin (x) + \frac{3}{4} = \frac{3}{2}, (0, 2 π)

2 cos (2 x) + sin (x) - 4 = 0

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

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