What is the general solution for solvefor θ,r= 7/(1+cos(θ)) ?

The general solution for solvefor θ,r= 7/(1+cos(θ)) is θ=arccos(7/r-1)+2pin,θ=-arccos(7/r-1)+2pin

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solvefor θ,r= 7/(1+cos(θ))

solve for

θ, r = \frac{7}{1 + cos ( θ )}

θ = arccos (\frac{7}{r} - 1) + 2 πn, θ = - arccos (\frac{7}{r} - 1) + 2 πn

Solution steps

r = \frac{7}{1 + cos ( θ )}

Switch sides

\frac{7}{1 + cos ( θ )} = r

Multiply both sides by

1 + cos (θ)

\frac{7}{1 + cos ( θ )} = r

Multiply both sides by

1 + cos (θ)

\frac{7}{1 + cos ( θ )} (1 + cos (θ)) = r (1 + cos (θ))

Simplify

7 = r (1 + cos (θ))

7 = r (1 + cos (θ))

Switch sides

r (1 + cos (θ)) = 7

Divide both sides by

r; r \neq = 0

r (1 + cos (θ)) = 7

Divide both sides by

r; r \neq = 0

\frac{r ( 1 + cos ( θ ) )}{r} = \frac{7}{r}; r \neq = 0

Simplify

1 + cos (θ) = \frac{7}{r}; r \neq = 0

1 + cos (θ) = \frac{7}{r}; r \neq = 0

Move

1

to the right side

1 + cos (θ) = \frac{7}{r}; r \neq = 0

Subtract

1

from both sides

1 + cos (θ) - 1 = \frac{7}{r} - 1; r \neq = 0

Simplify

cos (θ) = \frac{7}{r} - 1; r \neq = 0

cos (θ) = \frac{7}{r} - 1; r \neq = 0

Apply trig inverse properties

cos (θ) = \frac{7}{r} - 1

General solutions for

cos (θ) = \frac{7}{r} - 1

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

θ = arccos (\frac{7}{r} - 1) + 2 πn, θ = - arccos (\frac{7}{r} - 1) + 2 πn

θ = arccos (\frac{7}{r} - 1) + 2 πn, θ = - arccos (\frac{7}{r} - 1) + 2 πn

csc (θ) = - \frac{3}{5}, tan (2 θ)

1 - sin (x) = 0

x, sin (x) = e

x, sin (x) = 3

3 cos (2 θ) + 5 cos (θ) + 2 = 0

What is the general solution for solvefor θ,r= 7/(1+cos(θ)) ?
The general solution for solvefor θ,r= 7/(1+cos(θ)) is θ=arccos(7/r-1)+2pin,θ=-arccos(7/r-1)+2pin