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

The general solution for sin(x)=-1,0<x<2pi is x=(3pi)/2

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sin(x)=-1,0<x<2pi

sin (x) = - 1, 0 < x < 2 π

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

sin (x) = - 1, 0 < x < 2 π

General solutions for

sin (x) = - 1

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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 < x < 2 π

x = \frac{3 π}{2}

2 sin^{2} (x) - 7 sin (x) = 0

sec (x) = \frac{1}{3}

0 = 1 + 4 sin (θ)

x, u = cos (x^{2} y)

6 sin^{2} (x) - sin (x) cos (x) - 2 cos^{2} (x) = 0

What is the general solution for sin(x)=-1,0<x<2pi ?
The general solution for sin(x)=-1,0<x<2pi is x=(3pi)/2