How to find the tangent to a function passing through the origin

Question

My function is

$F(x)=Acos^{3}(x)+Bcos^2{x}+Ccos(x)+D$

and I need to find a function $y(x)=ax$ tangent to $F(x)$ and passing through the origin.

I tried analytically with $y=m(x-x_{0})+y_{0}$ but I got a huge mess and I don't know how to take it from there.

Is there any way to do it somehow easier?

Answer 1

Let us just consider the last two terms. The condition of tangency amounts to finding the extrema of

$$\frac{C\cos x+D}x.$$

The numerator of the derivative is

$$-C\sin(x)\,x-C\cos(x)-D.$$

This is a transcendental equation that cannot be solved analytically. There is no easy solution.

Answer 2

Yves Daoust provided a good answer, but let me expand a little bit. An approach to this problem is to parametrize the curve as $\langle x, F(x)\rangle$ in $\mathbb{R}^2$. Then, the tangent line to the curve is $$x\left< 1,\frac{dF}{dx}\right>+\left<x,F(x)\right>$$ Having the tangent line pass through the origin is equivalent to having the position vector $\left<x,F(x)\right>$ be a multiple of the tangent vector to the curve. Hence, the problem reduces to solving $$x\frac{dF}{dx}=F(x)$$ for $x$, and evaluating the derivative at that point. However, it does not seem to be analytically solvable, as was pointed out.