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Hello I am having some problems trying to solve the following trigonometric equation when is $$ -\tan(x) +3\sin(x) = \cos(x) $$ on the interval $0$ to $2\pi$.

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  • $\begingroup$ Yuck. Where does this problem come from? $\endgroup$ – user43208 Sep 14 '13 at 21:20
  • $\begingroup$ it came from one of my lecture notes $\endgroup$ – user94883 Sep 14 '13 at 21:42
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There are actually 4 real valued $x$ solutions to this equation on the interval $0 \leq x \leq 2 \pi$. Pulling something from trigonometry lecture notes and trying to solve it is good, but as it turns out not all trigonometric equations lend themselves to a solution by ordinary algebraic methods and trigonometric simplifications.

The correct and appropriate way to solve this equation is by numerical methods. You could either approximate the intercepts of $$f_1=-\tan(x)+3\sin(x) \,\, \operatorname{and } \,\, f_2=\cos(x),$$

or you could just approximate the zeroes of $$f(x)=-\tan(x)+3\sin(x)-\cos(x).$$

Graphically is good. Many calculators used in college math courses should do it. We have a host of numerical methods to do it, and that would probably deserve another post or modification to your question.

The short story is that this equality did not wind up in your lecture notes as an equation to solve for exact values, and if I am wrong I will happily eat mud just to see the magic.

Note that, using Mathematica, or say your Ti-89 calculator, or some other numeric solver, you will find 4 real roots on this interval at $\{0.492099, 1.17268, 3.3848, 5.0187\}$

As I think about this, for visual purposes, it looks nice and makes the most sense to solve this for $\tan(x)$, and then plot $f_1=\tan(x)$ and $f_2=3\sin(x)-\cos(x)$. That way you can see the asymptotes of the tangent function, and where the real valued intercepts are on the bound.

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