Trignometric Equation Solution Question : On the interval $[0,2\Pi]$ there is one point on the curve $r = \Theta  - 2cos\Theta$ whose x-coordinate is 2. Find the y-coordinate there.
The solution simply states: Solving $(\Theta - 2cos\Theta)cos\Theta = 2$ 
yields $\Theta \approx 5.201$ and so on....
I would like to know how it yields 5.201
My working: 
$(\Theta - 2cos\Theta)cos\Theta = 2$
Hence, either $cos\Theta = 2$ 
or, $(\Theta - 2cos\Theta)$ = 2
The former not being possible, we get:
$\Theta = 2(1+ cos\Theta)$
$\Theta/2 = 1 + cos\Theta$
$cos\Theta = (\Theta - 2)/2$
Hence $\Theta = arccos((\Theta - 2)/2)$
So, I don't understand how it results in $\Theta$ being 5.201
 A: Your method is wrong because you have assumed, in effect, the following:

if $ab=2$, then either $a=2$ or $b=2$.

This is not correct because there are actually many possibilities: you could have $a=4$, $b=\frac12$, or $a=-1000$, $b=-\frac1{500}$ and so on.  You are mixing things up with the following correct principle:

if $ab=0$, then either $a=0$ or $b=0$.

This is a special property of $0$, and it does not continue to be true if the $0$ is replaced by a different number.
As for the problem, so far as I can tell, the equation
$$(\theta-2\cos\theta)\cos\theta=2$$
is going to be very hard to solve exactly, and I expect they used an approximate method.  Such methods may or may not be part of the course you are studying.
A: Equations which mix polynomial and trigonometric terms do not show explicit solutions (it is already the case for $\theta=\cos(\theta)$ and only numerical methods can solve for the roots.
Let us consider the case of the function $$f(\theta)=\big(\theta - 2cos(\theta)\big)\cos(\theta)- 2$$ for which we want to find the first positive zero (in fact, there is an infinite number of solutions to $f(\theta)=0$). If you plot the function, you will notice that there is a root close to $2\pi$ (as Robert Israel commented).
For finding the root, let us use Newton method; this is one of the simplest root-finding method. Starting from a "reasonable" guess $\theta_0$, the method will update it accordind to $$\theta_{n+1}=\theta_{n}-\frac{f(\theta_{n})}{f'(\theta_{n})}$$ $$f'(\theta)=\sin (\theta ) (2 \cos (\theta )-\theta )+(2 \sin (\theta )+1) \cos (\theta )$$ So, let us use the method starting with $\theta_0=2\pi$; so, the successive generated iterates are $4$, $5.19942$, $5.20080$ which is the solution for six significant figures. 
If instead, as shown by the graph, you start iterating using $\theta_0=5$, the successive generated iterates are $5.18611$, $5.20071$, $5.20080$.
