# A complicated trigonometric equation

I have the following trigonometric equation $$f(\theta)=100(A_2 B_3 - A_3 B_2)^2 - (c_1B_3 - c_2 B_2)^2 - (c_2A_2 - c_1 A_3)^2=0,$$

where:
$A_2 = 3\cos(\theta)-5$
$B_2 = 3\sin(\theta)$
$A_3 = 3(\cos(\theta) - \sin(\theta))$
$B_3 = 3(\cos(\theta) + \sin(\theta))-6$
$c_1 = p_2^2 - 25 - A_2^2 - B_2^2$
$c_2 = -16 - A_3^2 - B_3^2$

and need to find all the values of $p_2$ for which $f(\theta)=0$ has 2, 4, 6 solutions in $[-\pi,\pi]$ and no solution. For example, if $p_2 = 4$, $f(\theta)=0$ has 2 solutions in $[-\pi,\pi]$. If $p_2 = 5$, $f(\theta)=0$ has 4 solutions. If $p_2 = 7$, $f(\theta)=0$ has 6 solutions. If $p_2 = 1$, $f(\theta)=0$ has no solution.These values follow from the graph of the function $f(\theta)$ (tested on MATLAB). So what if we want to generalize this approach? Any ideas?

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It took me a while to understand this question. Am I right in guessing that where it says "when" you mean "for which"? Also, do I understand correctly that you've already found the answer for $2$, $4$ and $6$ and are now looking only for the values of $p_2$ for which $f(\theta)$ has no solution? (Where you're presumably only counting real solutions in $[0,2\pi)$?) – joriki Dec 20 '11 at 21:49
Thanks for the clarification. I suspect you mean $(-\pi,\pi]$? Otherwise you'd be counting $-\pi$ and $\pi$ as separate solutions. – joriki Dec 21 '11 at 1:24
Hi! I apologize for any inconvenience. Firstly, I forgot to mention the interval in which the equation has solutions. I edited (did my best!) my question so that you can understand. Secondly, I've found some arbitrary $p_2$ for which the $f(\theta)$ has 2,4,6 solutions in $[-\pi, \pi]$ and no solution. The goal is to find all the values of $p_2$. – user21530 Dec 21 '11 at 1:39
I thank you. No, I mean $[-\pi, \pi]$. As one can see, $\pi$ and $-\pi$ are not solutions of the equation. – user21530 Dec 21 '11 at 1:48
Then it seems you mean "all real values of $p_2$"? There are complex values of $p_2$ for which $\theta=\pm\pi$ is a solution. Also, the current wording does not express that you found some arbitrary $p_2$ and not all values. "If ... then ..." is an implication. From your comment, it appears that you mean e.g. "If $p_2=4$, then $f(\theta)=0$ has $2$ solutions in $[\pi,-\pi]$"? Further, note that $A_3$ and $B_3$ can be simplified, since $\cos\pi/4=\sin\pi/4=1/\sqrt2$. – joriki Dec 21 '11 at 10:58