What is the general solution for 2asin(x)+a=0 ?

The general solution for 2asin(x)+a=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2asin(x)+a=0

2 a sin (x) + a = 0

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

2 a sin (x) + a = 0

Move

a

to the right side

2 a sin (x) + a = 0

Subtract

a

from both sides

2 a sin (x) + a - a = 0 - a

Simplify

2 a sin (x) = - a

2 a sin (x) = - a

Divide both sides by

2 a; a \neq = 0

2 a sin (x) = - a

Divide both sides by

2 a; a \neq = 0

\frac{2 a sin ( x )}{2 a} = \frac{- a}{2 a}; a \neq = 0

Simplify

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

Simplify

\frac{2 a sin ( x )}{2 a} : sin (x)

\frac{2 a sin ( x )}{2 a}

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

a

= sin (x)

Simplify

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

\frac{- a}{2 a}

Apply the fraction rule:

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

= - \frac{a}{2 a}

Cancel the common factor:

a

= - \frac{1}{2}

sin (x) = - \frac{1}{2}; a \neq = 0

sin (x) = - \frac{1}{2}; a \neq = 0

sin (x) = - \frac{1}{2}; a \neq = 0

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

What is the general solution for 2asin(x)+a=0 ?
The general solution for 2asin(x)+a=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin