I'm facing the problem of solving $$\sin(x)+x \cdot \cos(x)=0$$ using $$\tan(x)=\sin(x)/\cos(x)$$ I end at $$x+\tan(x)=0$$ on the other hand, I also tried $\cos(x)= \pm \sqrt{1-\sin^2(x)}$ resulting in $x^2-x^2\cdot \sin^2(x)-\sin^2(x)=0$ OR using $\sin(x)= \pm \sqrt{1-\cos^2(x)}$ I get $1-\cos^2(x)-x^2\cdot \cos^2(x)=0$
In all these cases I have no idea on how to get to a solution like $x=...$. As far is I can see, I should try getting $x$ into the trigonometric functions, but I don't see how I could?
tanks for all your help
$\endgroup$2 Answers
$\begingroup$Generally, trig functions don't talk to polynomials, so you're not going to get an algebraic solution to this one. Your favorite numeric solver will have no problem. If you draw the graphs of $y=\tan x$ and $y=-x$ you'll have an easy time finding where to look for roots.
$\endgroup$ 1 $\begingroup$$$\sin{x} + x\cos{x} = 0$$
$$\cos{x} (\tan{x} + x) =0$$
This would have been true if $\cos{x}=0$ but note that in this case the $\tan{x}$ term blows up and the equation is $0\times\infty$. The other possibility is when $\tan{x}=-x$ which has an infinite number of solutions. That $x=0$ is a solution is obvious, the others can be found graphically/numerically.
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