What is the general solution for r=(2a)/((1+cos(x))) ?

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

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r=(2a)/((1+cos(x)))

r = \frac{2 a}{( 1 + cos ( x ) )}

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

Solution steps

r = \frac{2 a}{( 1 + cos ( x ) )}

Switch sides

\frac{2 a}{1 + cos ( x )} = r

Multiply both sides by

1 + cos (x)

\frac{2 a}{1 + cos ( x )} = r

Multiply both sides by

1 + cos (x)

\frac{2 a}{1 + cos ( x )} (1 + cos (x)) = r (1 + cos (x))

Simplify

2 a = r (1 + cos (x))

2 a = r (1 + cos (x))

Switch sides

r (1 + cos (x)) = 2 a

Divide both sides by

r; r \neq = 0

r (1 + cos (x)) = 2 a

Divide both sides by

r; r \neq = 0

\frac{r ( 1 + cos ( x ) )}{r} = \frac{2 a}{r}; r \neq = 0

Simplify

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

Move

1

to the right side

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

Subtract

1

from both sides

1 + cos (x) - 1 = \frac{2 a}{r} - 1; r \neq = 0

Simplify

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

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

Apply trig inverse properties

cos (x) = \frac{2 a}{r} - 1

General solutions for

cos (x) = \frac{2 a}{r} - 1

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

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

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

cos^{3} (x) + cos^{2} (x) + cos (x) + 1 = 0

\frac{cos ^{4} ( x )}{3} = sin^{2} (x)

sin^{3} (x) cos (x) - sin^{2} (x) = 0

sin^{2} (x) + sin^{4} (x) = 0

cos^{2} (4 5^{\circ} - a) - sin^{2} (4 5^{\circ} - a) = sin^{2} (a)

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