Here's the question you clicked on:
kb6
sin (2x) =-1/2
\(\bf sin (2x) =-\cfrac{1}{2}\quad \textit{take }sin^{-1} \\ \quad \\ sin^{-1}[sin ({\color{red}{ 2x}})]=sin^{-1}\left(-\frac{1}{2}\right)\implies {\color{red}{ 2x}}=sin^{-1}\left(-\frac{1}{2}\right)\) solve for "x"
It says to solve the equation on the interval 0<x<2pi
yes so... what did you get for "x"?
\[\sin 2x=-\frac{ 1 }{ 2 }=-\sin \frac{ \pi }{6 }=\sin \left( \pi+\frac{ \pi }{6 } \right),\sin \left( 2\pi-\frac{ \pi }{ 6 } \right),\sin \left( 2\pi+\pi+\frac{ \pi }{ 6 } \right),\] \[\sin \left( 2\pi+2\pi-\frac{ \pi }{6 } \right)\] 2x=? x=?