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

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

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sin(θ)=-(sqrt(3))/2 ,pi<θ<(3pi)/2

sin (θ) = - \frac{3}{2}, π < θ < \frac{3 π}{2}

θ = \frac{4 π}{3}

Degrees

θ = 24 0^{\circ}

Solution steps

sin (θ) = - \frac{3}{2}, π < θ < \frac{3 π}{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

Solutions for the range

π < θ < \frac{3 π}{2}

θ = \frac{4 π}{3}

cos (3 x) = sin (x - 3 0^{\circ})

4 sin (θ) = 4 cos (2 θ)

tan (5 θ + 30) = 1

885 cos (θ) - 70 = cos (25 0^{\circ})

tan (x) + 5 = 0, x \leq 0 < 2 π

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