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

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

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

2 sin (x) = - 1, 0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Degrees

x = 21 0^{\circ}, x = 33 0^{\circ}

Solution steps

2 sin (x) = - 1, 0 \leq x < 2 π

Divide both sides by

2

2 sin (x) = - 1

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 1}{2}

Simplify

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \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}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

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