Question

Find at least one solution to the following equation:
sin(x^2 − 1)/1 − sin(x^2 − 1) = sin(x) + sin^2(x) + sin^3(x) + sin^4(x) + · · ·

Answer 1

Right side of the equation is a geometric series with $a = sin(x)$ and $r = \frac{sin^2(x)}{sin(x)} = sin(x)$.

The sum of a geometric series:

$$S = \frac{a}{1-r}$$ $$S = \frac{sin(x)}{1-sin(x)}$$

Hence the equation is:

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

$$sin(x^2-1)(1-sin(x)) = sin(x)(1-sin(x^2-1))$$

$$sin(x^2-1) = sin(x)$$

Since the period the function $y = sin(x)$ is $2\pi$, the equation is:

$$x^2-1 = x +2\pi n$$

$$x^2-x-(1+2\pi n) = 0$$

$$x_1 = \frac{1-\sqrt{5+8\pi n }}{2}, x_2 = \frac{1+\sqrt{5+8\pi n }}{2}, n = 0, 1, 2, ...$$

In particular, for n = 0 we have:

$$x_1 = \frac{1-\sqrt{5}}{2}, x_2 = \frac{1+\sqrt{5}}{2}$$

For n = 1:

$$x_1 = \frac{1-\sqrt{5+8\pi}}{2}, x_2 = \frac{1+\sqrt{5+8\pi}}{2}$$