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Popular >

solvefor θ,r=(10)/(3+2cos(θ))

Solution

solve for

θ, r = \frac{10}{3 + 2 cos ( θ )}

Solution

θ = arccos (\frac{10 - 3 r}{2 r}) + 2 πn, θ = - arccos (\frac{10 - 3 r}{2 r}) + 2 πn

Solution steps

r = \frac{10}{3 + 2 cos ( θ )}

Switch sides

\frac{10}{3 + 2 cos ( θ )} = r

Multiply both sides by

3 + 2 cos (θ)

\frac{10}{3 + 2 cos ( θ )} = r

Multiply both sides by

3 + 2 cos (θ)

\frac{10}{3 + 2 cos ( θ )} (3 + 2 cos (θ)) = r (3 + 2 cos (θ))

Simplify

10 = r (3 + 2 cos (θ))

10 = r (3 + 2 cos (θ))

Switch sides

r (3 + 2 cos (θ)) = 10

Divide both sides by

r; r \neq = 0

r (3 + 2 cos (θ)) = 10

Divide both sides by

r; r \neq = 0

\frac{r ( 3 + 2 cos ( θ ) )}{r} = \frac{10}{r}; r \neq = 0

Simplify

3 + 2 cos (θ) = \frac{10}{r}; r \neq = 0

3 + 2 cos (θ) = \frac{10}{r}; r \neq = 0

Move

3

to the right side

3 + 2 cos (θ) = \frac{10}{r}; r \neq = 0

Subtract

3

from both sides

3 + 2 cos (θ) - 3 = \frac{10}{r} - 3; r \neq = 0

Simplify

2 cos (θ) = \frac{10}{r} - 3; r \neq = 0

2 cos (θ) = \frac{10}{r} - 3; r \neq = 0

Divide both sides by

2; r \neq = 0

2 cos (θ) = \frac{10}{r} - 3; r \neq = 0

Divide both sides by

2; r \neq = 0

\frac{2 cos ( θ )}{2} = \frac{\frac{10}{r}}{2} - \frac{3}{2}; r \neq = 0

Simplify

\frac{2 cos ( θ )}{2} = \frac{\frac{10}{r}}{2} - \frac{3}{2}

Simplify

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos (θ)

Simplify

\frac{\frac{10}{r}}{2} - \frac{3}{2} : \frac{10 - 3 r}{2 r}

\frac{\frac{10}{r}}{2} - \frac{3}{2}

Apply rule

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

= \frac{\frac{10}{r} - 3}{2}

Join

\frac{10}{r} - 3 : \frac{10 - 3 r}{r}

\frac{10}{r} - 3

Convert element to fraction:

3 = \frac{3 r}{r}

= \frac{10}{r} - \frac{3 r}{r}

Since the denominators are equal, combine the fractions:

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

= \frac{10 - 3 r}{r}

= \frac{\frac{10 - 3 r}{r}}{2}

Apply the fraction rule:

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

= \frac{10 - 3 r}{r \cdot 2}

cos (θ) = \frac{10 - 3 r}{2 r}; r \neq = 0

cos (θ) = \frac{10 - 3 r}{2 r}; r \neq = 0

cos (θ) = \frac{10 - 3 r}{2 r}; r \neq = 0

Apply trig inverse properties

cos (θ) = \frac{10 - 3 r}{2 r}

General solutions for

cos (θ) = \frac{10 - 3 r}{2 r}

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

θ = arccos (\frac{10 - 3 r}{2 r}) + 2 πn, θ = - arccos (\frac{10 - 3 r}{2 r}) + 2 πn

θ = arccos (\frac{10 - 3 r}{2 r}) + 2 πn, θ = - arccos (\frac{10 - 3 r}{2 r}) + 2 πn

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