What is the general solution for solvefor x,(-y)/2+sin(x)=0 ?

The general solution for solvefor x,(-y)/2+sin(x)=0 is x=arcsin(y/2)+2pin,x=pi+arcsin(-y/2)+2pin

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solvefor x,(-y)/2+sin(x)=0

solve for

x, \frac{- y}{2} + sin (x) = 0

x = arcsin (\frac{y}{2}) + 2 πn, x = π + arcsin (- \frac{y}{2}) + 2 πn

Solution steps

\frac{- y}{2} + sin (x) = 0

Move

\frac{- y}{2}

to the right side

\frac{- y}{2} + sin (x) = 0

Subtract

\frac{- y}{2}

from both sides

\frac{- y}{2} + sin (x) - \frac{- y}{2} = 0 - \frac{- y}{2}

Simplify

\frac{- y}{2} + sin (x) - \frac{- y}{2} = 0 - \frac{- y}{2}

Simplify

\frac{- y}{2} + sin (x) - \frac{- y}{2} : sin (x)

\frac{- y}{2} + sin (x) - \frac{- y}{2}

Add similar elements:

\frac{- y}{2} - \frac{- y}{2} = 0

= sin (x)

Simplify

0 - \frac{- y}{2} : \frac{y}{2}

0 - \frac{- y}{2}

Apply the fraction rule:

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

= 0 - (- \frac{y}{2})

Apply rule

- (- a) = a

= 0 + \frac{y}{2}

0 + \frac{y}{2} = \frac{y}{2}

= \frac{y}{2}

sin (x) = \frac{y}{2}

sin (x) = \frac{y}{2}

sin (x) = \frac{y}{2}

Apply trig inverse properties

sin (x) = \frac{y}{2}

General solutions for

sin (x) = \frac{y}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{y}{2}) + 2 πn, x = π + arcsin (- \frac{y}{2}) + 2 πn

x = arcsin (\frac{y}{2}) + 2 πn, x = π + arcsin (- \frac{y}{2}) + 2 πn

cos (a) = 0.33

tan (θ) = \frac{3}{- 2}

4 sin^{2} (r) - sin (r) = 0

tan (θ) = \frac{3}{- 3}

cos (a) = 0.24

What is the general solution for solvefor x,(-y)/2+sin(x)=0 ?
The general solution for solvefor x,(-y)/2+sin(x)=0 is x=arcsin(y/2)+2pin,x=pi+arcsin(-y/2)+2pin