Question

Find at least one solution to the following equation:

Answer 1

For |x| < 1, we have

$1+x+x^2+\cdots=\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}$

We have |sin(x)| ≤ 1, with equality when x = ± π/2. For either of these, the right side of the equation does not converge, since it's either an infinite sum of 1s or an infinite alternating sum of 1 and -1.

So for |sin(x)| < 1, we have

$\sin(x)+\sin^2(x)+\sin^3(x)+\cdots=\dfrac1{1-\sin(x)}-1=\dfrac{\sin(x)}{1-\sin(x)}$

so the equation is equivalent to

$\dfrac{\sin(x^2-1)}{1-\sin(x^2-1)}=\dfrac{\sin(x)}{1-\sin(x)}$

One obvious set of solutions occurs when x = x² - 1 :

x² - x - 1 = 0   →   x = (1 ± √(5))/2

i.e. the golden ratio φ ≈ 1.618 and 1 - φ ≈ 0.618.