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

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

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(sqrt(2))/2 =sin(θ)

\frac{2}{2} = sin (θ)

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

Degrees

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

Solution steps

\frac{2}{2} = sin (θ)

Switch sides

sin (θ) = \frac{2}{2}

General solutions for

sin (θ) = \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}

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

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

cos (a) = 1, 8 b = 1, 5 c = 23

sin (θ) cos (θ) = \frac{2 3}{5}

tan (2 5^{\circ}) = cot (2 x + 5)

sin (a) = 0.4

sin (a) = 0.2

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