Trignometric Equation Solution

Question

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

Answer 1

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 $5.18611$, $5.20071$, $5.20080$.

Answer 2

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.

Answer 3

As to the zooming method, suppose that you found the interval $[5.1,5.3]$ as likely containing a root. Compute the table of values of $f(x)=(x-2\cos x)\cos x -2$ as $$ \begin{array}{c|l|l} k & x_k & f(x_k)\\ \hline 1 & 5.1 &-0.358047860136601\\ 2 & 5.3 & 0.323522172521382\\ \end{array} $$ so there is a definite sign change. By the secant method compute $$ x_3=\frac{x_1f(x_2)-x_2f(x_1)}{f(x_2)-f(x_1)}=5.20506561115672 $$ and compute the new function table. \begin{array}{c|l|l} k & x_k & f(x_k)\\ \hline 1 & 5.1 &-0.358047860136601\\ 3 & 5.20506561115672 & 0.0144912491098061 \\ 2 & 5.3 & 0.323522172521382\\ \end{array} Compute $x_4$ from the secant over the interval $[x_2,x_3]$ $$ x_4=\frac{x_2f(x_3)-x_3f(x_2)}{f(x_3)-f(x_2)}=5.20061389524872 $$ and repeat … \begin{array}{c|c|c} k & x_k & f(x_k)\\ \hline 1 & 5.1 &-0.358047860136601\\ 4 & 5.20061389524872 &-0.000638165265964163 \\ 7 & 5.20080132814502 & 0.000000000000002220 \\ 6 & 5.20080132817247 & 0.000000000093464347 \\ 5 & 5.20080167055819 & 0.0000011657432845702 \\ 3 & 5.20506561115672 & 0.0144912491098061 \\ 2 & 5.3 & 0.323522172521382\\ \end{array} As per the supplied solution, the accuracy of $x_4$ should already be sufficient.