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

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 $X^2+Y^2=1$.

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.
