What is the general solution for solvefor θ,r=2cos(θ) ?

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

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solvefor θ,r=2cos(θ)

solve for

θ, r = 2 cos (θ)

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

Solution steps

r = 2 cos (θ)

Switch sides

2 cos (θ) = r

Divide both sides by

2

2 cos (θ) = r

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{r}{2}

Simplify

cos (θ) = \frac{r}{2}

cos (θ) = \frac{r}{2}

Apply trig inverse properties

cos (θ) = \frac{r}{2}

General solutions for

cos (θ) = \frac{r}{2}

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

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

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

sin (2 x) = \frac{2 2}{3}

arccos (x) = \frac{15}{17}

4 tan^{2} (θ) - 9 = 0

sin (θ) = 7.19

3 = \frac{π}{3} sec^{2} (c)

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