If $S=1+\dfrac{1}{\sin x}+\dfrac{1}{\sin^2x}+...+\dfrac{1}{\sin^n x}+...=\dfrac23,$ what's the measure of angle $x$?
We have an infinite geometric series with first term $a_1=1$ and common ratio $q=\dfrac{1}{\sin x}$. Then $$S=\dfrac{1}{1-\frac{1}{\sin x}}=\dfrac{\sin x}{\sin x-1}=\dfrac{2}{3}\iff3\sin x=2\sin x-2\iff\sin x=-2, $$ which has no solutions. Am I making a silly mistake somewhere?
This is a divergent serie.$|\sin(x)|\leq 1$ so the reciprocal is greater or equal to 1.
If you consider for example $\sin(x)=\dfrac{1}{2}$ you have $\dfrac{1}{\sin(x)}=2$
And the series became: $1+2+2^2+2^3+2^4+...\to\infty$
Assuming that the geometric series $$ S(x)=1+\sum_{n=1}^{\infty}(\sin x)^{-n} \tag{1} $$ converges, the OP showed that it equals $\frac{2}{3}$ if $\sin x=-2$. On the other hand, as already explained by others, the series diverges if $x$ is real, for $|\sin x|\leq 1$. Now, although $\sin x=-2$ has no real solution, it has complex solutions. Indeed, if we write $x=u+iv$, where $u$ and $v$ are real, then$^{(*)}$ $$ \sin x = \sin(u+iv) = \sin u\cosh v + i\cos u \sinh v, \tag{2} $$ hence $$ \sin x = -2 \iff \begin{cases} \sin u\cosh v = -2, \\ \cos u\sinh v = 0. \end{cases} \tag{3} $$ One can easily check that $(3)$ has the following solutions: \begin{align} u&=-\frac{\pi}{2}+2n\pi\quad(n=0,\pm1,\pm2,\ldots), \\ v&=\cosh^{-1}(2)=\pm\ln(2+\sqrt{3}), \tag{4} \end{align} therefore $S(x)=\frac{2}{3}$ at the "angles" $x=(-\frac{\pi}{2}+2n\pi)\pm i\ln(2+\sqrt{3})$ $(n=0,\pm1,\pm2,\ldots)$.
$^{(*)}$https://en.wikipedia.org/wiki/Trigonometric_functions#In_the_complex_plane
