# Solve tan(x)+cos(x)=1/2

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):

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

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

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

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

• 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 – Dr. Mathva Mar 19 at 21:08
• Of course it is possible (cf. Ferrari's formula for the quartic, for example). – Allawonder Mar 19 at 21:19
• @Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics – pasaba por aqui Mar 19 at 21:23
• @pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was. – Allawonder Mar 19 at 21:26
• @Allawonder: yes, your comment is correct, I only adding context – 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 $$X^2+Y^2=1$$.
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 $$3y^4+4y^3-4y^2+4y+1=0$$