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

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

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(sqrt(3))/2 =sin(θ)

\frac{3}{2} = sin (θ)

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

Degrees

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

Solution steps

\frac{3}{2} = sin (θ)

Switch sides

sin (θ) = \frac{3}{2}

General solutions for

sin (θ) = \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}

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

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

cos (\frac{π}{2} - θ) = cos (θ)

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

cos^{2} (θ) + cos (θ) = 0

sin (θ) = \frac{3}{2}

sin^{2} (θ) + 2 cos^{2} (θ) = 2

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