Question

2Sinx+1=0 Use n as an integer.

Answer 1

apparently 7pi/6 + 2npi isn't the answer

Answer 2

\(\dfrac{7\pi}{6}+2\pi n\) is correct answer, but you are missing something.
\[2\sin x+1=0\Longrightarrow\sin x=-\dfrac{1}{2}\]There is two points on unit circle in range \([0,2\pi)\) those satisfy this equation.