What is the general solution for r=a(1-sin(x)) ?

The general solution for r=a(1-sin(x)) is x=arcsin(-(r-a)/a)+2pin,x=pi+arcsin((r-a)/a)+2pin

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Popular >

r=a(1-sin(x))

Solution

r = a (1 - sin (x))

Solution

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

Solution steps

r = a (1 - sin (x))

Switch sides

a (1 - sin (x)) = r

Divide both sides by

a; a \neq = 0

a (1 - sin (x)) = r

Divide both sides by

a; a \neq = 0

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

Simplify

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Divide both sides by

- 1; a \neq = 0

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

Divide both sides by

- 1; a \neq = 0

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

Simplify

\frac{- sin ( x )}{- 1} = \frac{\frac{r}{a}}{- 1} - \frac{1}{- 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{r}{a}}{- 1} - \frac{1}{- 1} : - \frac{r - a}{a}

\frac{\frac{r}{a}}{- 1} - \frac{1}{- 1}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{\frac{r}{a} - 1}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{r}{a} - 1}{1}

Join

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

\frac{r}{a} - 1

Convert element to fraction:

1 = \frac{1 a}{a}

= \frac{r}{a} - \frac{1 \cdot a}{a}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{r - 1 \cdot a}{a}

Multiply:

1 \cdot a = a

= \frac{r - a}{a}

= - \frac{\frac{r - a}{a}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= - \frac{r - a}{a}

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for r=a(1-sin(x)) ?
The general solution for r=a(1-sin(x)) is x=arcsin(-(r-a)/a)+2pin,x=pi+arcsin((r-a)/a)+2pin