What is the general solution for -sin(x)= 1/2 ?

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

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-sin(x)= 1/2

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

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

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Divide both sides by

- 1

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

Divide both sides by

- 1

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

Simplify

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

Simplify

\frac{- sin ( x )}{- 1} : sin (x)

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin (x)

Simplify

\frac{\frac{1}{2}}{- 1} : - \frac{1}{2}

\frac{\frac{1}{2}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{1}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{1}{2}}{1} = \frac{1}{2}

= - \frac{1}{2}

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

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

sin (x) sec (x) = 2 sin (x), 0 \leq x \leq 2 π

3 cos^{2} (y) - 1 = cos (y) + cos^{2} (y), π \leq y \leq 2 π

sec^{2} (x) + tan (x) = 1, 0 \leq x \leq 2 π

6 tanh^{2} (x) + 5 sech (x) - 7 = 0

2 sin (x) + 2 = 0, 0 \leq x \leq 2 π

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