Finding the cubic near minimax approximation for sin(x) on (0,pi/2)

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I am really stuck here. Here is the question that I have.

Find the cubic near minimax approximation for $f(x)=\sin(x)$ on $(0,\pi/2)$.

So I defined $h(x)=ax^3 + bx^2 + cx + d - sin(x)$

The max of this occurs at $dh/dx=0$, so that means

$$\frac{dh}{dx} = 3ax^2 + 2bx + c - \cos(x) = 0.$$

Now I am stuck. I wanted to solve this for $x$, and then find $h(0), \; h(\text{whatever } x \text{ gives me from the equation above})$, and $h(\pi/2)$. I would then use these solutions to solve for my coefficients. However, I cannot solve $dh/dx = 0$ for $x$. Not even wolfram can do it, it is too computationally intensive. Can someone please help me?

Thank you for your time!