Trigonometric Equations

To solve the given series equation:$1+ \sin x+ \sin^2 x+ \sin^3 x+ \dots = 4+2\sqrt{3}$we recognize that this is an infinite geometric series with the first term $a=1$ and the common ratio $r= \sin x$. The sum of an infinite geometric series is given by:$S=\frac{a}{1-r}$Here, substituting the given values:$\frac{1}{1- \sin x}=4+2\sqrt{3}$To find $\sin x$, we solve:$1- \sin x=\frac{1}{4+2\sqrt{3}}$Rationalizing the denominator gives:$1- \sin x=\frac{4-2\sqrt{3}}{4}=1-\frac{\sqrt{3}}{2}$Thus:$\sin x=\frac{\sqrt{3}}{2}$Given the constraints $0< x< \pi$ and $x \neq \frac{\pi}{2}$, we find that $x$ could be $\frac{\pi}{3}$ or $\frac{2\pi}{3}$. The solutions within this range are:$x=\frac{\pi}{3}, \frac{2\pi}{3}$