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

The general solution for 2sin(x)=-1,-pi<= x<= pi is x=-(5pi)/6 ,x=-pi/6

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

2 sin (x) = - 1, - π \leq x \leq π

x = - \frac{5 π}{6}, x = - \frac{π}{6}

Degrees

x = - 15 0^{\circ}, x = - 3 0^{\circ}

Solution steps

2 sin (x) = - 1, - π \leq x \leq 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

- π \leq x \leq π

x = - \frac{5 π}{6}, x = - \frac{π}{6}

cot (- \frac{π}{5}) - sec (x) = 1.5

\frac{1}{cos ^{2} ( x )} + (\frac{1}{2 cos ( x )}) = 0

3 sin (x) + cos (x) = 2

cos (α) = \frac{1 + \frac{1}{5}}{2}

(tan (x)) (sin^{2} (x) - a) (sec (x) - a) = 0, 0 < a < 1

What is the general solution for 2sin(x)=-1,-pi<= x<= pi ?
The general solution for 2sin(x)=-1,-pi<= x<= pi is x=-(5pi)/6 ,x=-pi/6