Is it possible (not numerically) to find the $x$ such as:

$$ tan(x)+cos(x)=1/2 $$


All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):

$$ \frac{\sqrt{1-c^2}}{c}+c=\frac{1}{2} $$

$$ \sqrt{1-c^2}+c^2=\frac{1}{2}c $$

$$ 1-c^2=c^2(\frac{1}{2}-c)^2=c^2(\frac{1}{4}-c+c^2) $$

$$ c^4-c^3+\frac{5}{4}c^2-1=0 $$

  • $\begingroup$ Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2 $\endgroup$ – Dr. Mathva Mar 19 at 21:08
  • $\begingroup$ Of course it is possible (cf. Ferrari's formula for the quartic, for example). $\endgroup$ – Allawonder Mar 19 at 21:19
  • $\begingroup$ @Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics $\endgroup$ – pasaba por aqui Mar 19 at 21:23
  • $\begingroup$ @pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was. $\endgroup$ – Allawonder Mar 19 at 21:26
  • $\begingroup$ @Allawonder: yes, your comment is correct, I only adding context $\endgroup$ – pasaba por aqui Mar 19 at 21:29

If we set $X=\cos x$ and $Y=\sin x$, the equation becomes $$ Y=\frac{1}{2}X-X^2 $$ so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.

enter image description here

This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.

The equation becomes $$ X^4-X^3+\frac{5}{4}X^2-1=0 $$ as you found out. The two real roots are approximately $$ -0.654665139167 \qquad 0.921490878816 $$ These correspond to $x=\pm2.284535877184578$ and $x=\pm0.39889463967156$, that correspond to what WolframAlpha finds.


If we put $t=x/2$ then we get $${2\tan t\over 1-\tan ^2t} +2\cos ^2 t -1={1\over 2}$$

Let $y= \tan t$. Since $\cos ^2t = {1\over 1+y^2}$ we get $$3y^4+4y^3-4y^2+4y+1=0$$

which I'm not sure if any helps. :(


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