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

The general solution for sin(θ)=(-sqrt(3))/2 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin

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sin(θ)=(-sqrt(3))/2

sin (θ) = \frac{- 3}{2}

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = \frac{- 3}{2}

Apply the fraction rule:

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

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{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 cos (\frac{x}{5}) - 2 = 0

2 sin^{2} (x) + 3 = 7 sin (x)

4 sin (x) = cos (x)

25 sin^{2} (θ) - 1 = 0

2 cos (x) sin (x) + 9 sin (x) = 0

What is the general solution for sin(θ)=(-sqrt(3))/2 ?
The general solution for sin(θ)=(-sqrt(3))/2 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin