What is the general solution for -1/2 =sin(θ) ?

The general solution for -1/2 =sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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-1/2 =sin(θ)

- \frac{1}{2} = sin (θ)

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- \frac{1}{2} = sin (θ)

Switch sides

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

2 sin (2 x) + 3 = 2

csc (θ) = 11

4 sin^{4} (θ) - 7 sin^{2} (θ) + 3 = 0

cos (v) = - 0.995

5 sin (X) - 2 = 2 sin (X)

What is the general solution for -1/2 =sin(θ) ?
The general solution for -1/2 =sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin