What is the general solution for r=a(1+cos(x)) ?

The general solution for r=a(1+cos(x)) is x=arccos(r/a-1)+2pin,x=-arccos(r/a-1)+2pin

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r=a(1+cos(x))

r = a (1 + cos (x))

x = arccos (\frac{r}{a} - 1) + 2 πn, x = - arccos (\frac{r}{a} - 1) + 2 πn

Solution steps

r = a (1 + cos (x))

Switch sides

a (1 + cos (x)) = r

Divide both sides by

a; a \neq = 0

a (1 + cos (x)) = r

Divide both sides by

a; a \neq = 0

\frac{a ( 1 + cos ( x ) )}{a} = \frac{r}{a}; a \neq = 0

Simplify

1 + cos (x) = \frac{r}{a}; a \neq = 0

1 + cos (x) = \frac{r}{a}; a \neq = 0

Move

1

to the right side

1 + cos (x) = \frac{r}{a}; a \neq = 0

Subtract

1

from both sides

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{r}{a} - 1) + 2 πn, x = - arccos (\frac{r}{a} - 1) + 2 πn

x = arccos (\frac{r}{a} - 1) + 2 πn, x = - arccos (\frac{r}{a} - 1) + 2 πn

4 sin^{2} (x) + 2 = 3

2 cos^{2} (x) + 7 sin (x) + 2 = 0

cos^{2} (x) = 0.5

sin (x) - cos (x) = - 1

4 cos (θ) + 22 = 0

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