How to solve equations like these?

Question

If $f(x) = x^{2} - \frac{\cos x}{2}$ and $g(x) = x\frac{\sin x}{2}$. Find the points at which $f(x) = g(x)$.

I'm stumped and have no idea how to proceed or solve questions like these.

Answer 1

As pointed out, it is impossible, or difficult, to find an analytic solution. You can use NR to approximate $f(x)=0$ for $f(x)=2x^2-x\sin x-\cos x$: $$\begin{align} x_{n+1}&=x_n-\frac{f(x_n)}{f'(x_n)}\\ &=x_n-\frac{2x_n^2-x_n\sin x_n-\cos x_n}{4x_n-\cos x_n}\\ &=\frac{(x_n^2-1)\cos x_n-x_n(\sin x_n+2x)}{x_n(\cos x_n-4)} \end{align}$$ $$$$ Thus with sufficient iterations, you get $x=0.7101850989$, which gives $f(x)=0$.