Simply Nice

You should specify atleast $3$ terms.

Always remember, two terms doesn't refer to an unique series ( or sequence ).

That's we've fundamental properties of well known sequences like A.P in terms of three terms :)

There are many ways to solve this problem. But the way that I really like is to use the below identity to get the series telescoped.

$\huge \boxed{\color{#D61F06}{\sec^{2}(x)\,=\,4 \cdot \csc^{2}(2x) - \csc^{2}(x)}}$ .

Using the identity $\sec^2x \; =\; 4\mathrm{cosec}^2 2x - \mathrm{cosec}^2x$ we see that $\begin{array}{rcl} \displaystyle \sum_{n=1}^N \frac{\sec^2\left(\frac{x}{2^n}\right)}{4^n} & = & \displaystyle \sum_{n=1}^N \frac{4\mathrm{cosec}^2\left(\frac{x}{2^{n-1}}\right) - \mathrm{cosec}^2\left(\frac{x}{2^n}\right)}{4^n} \\ & = & \displaystyle \sum_{n=1}^N \left\{ \frac{\mathrm{cosec}^2\left(\frac{x}{2^{n-1}}\right)}{4^{n-1}} - \frac{\mathrm{cosec}^2\left(\frac{x}{2^{n}}\right)}{4^{n}} \right\} \\ & = & \displaystyle \mathrm{cosec}^2x - \frac{\mathrm{cosec}^2\left(\frac{x}{2^{N}}\right)}{4^{N}} \\ & = & \displaystyle \mathrm{cosec}^2x - x^{-2} \left(\frac{\frac{x}{2^N}}{\sin\big(\frac{x}{2^N}\big)}\right)^2 \end{array}$ Letting $N \to \infty$ , we deduce that $\xi(x) \; = \; \mathrm{cosec}^2x - x^{-2}$ The sketch below shows that there are $\boxed{2}$ solutions of the equation $\xi(x) = 2$ in the range $0 < x < 5$ .