What is the general solution for sin(θ)=(-1)/2 ,0<= θ<= 4pi ?

The general solution for sin(θ)=(-1)/2 ,0<= θ<= 4pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ=(19pi)/6 ,θ=(23pi)/6

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sin(θ)=(-1)/2 ,0<= θ<= 4pi

sin (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{19 π}{6}, θ = \frac{23 π}{6}

Degrees

θ = 21 0^{\circ}, θ = 33 0^{\circ}, θ = 57 0^{\circ}, θ = 69 0^{\circ}

Solution steps

sin (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

Solutions for the range

0 \leq θ \leq 4 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{19 π}{6}, θ = \frac{23 π}{6}

4 cos^{2} (x) + 2 cos (x) = 2, 0 \leq x \leq 2 π

arctan (x) + arctan (2 x) = \frac{π}{4}

1 + cos (θ) = \frac{2 + 3}{2}

1 + 3 sin (x) = 2

cos (\frac{x}{2} + \frac{π}{3}) = \frac{1}{2}

What is the general solution for sin(θ)=(-1)/2 ,0<= θ<= 4pi ?
The general solution for sin(θ)=(-1)/2 ,0<= θ<= 4pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ=(19pi)/6 ,θ=(23pi)/6