3
$\begingroup$

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):

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

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

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

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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. :(

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.