Question

solve the equation on the interval 02pi 2sin theta +1=0

Answer 1

\[2\sin \theta+1=0\]

Answer 2

\[\sin \theta=\frac{ -1 }{ 2 }=-\sin \frac{ \pi }{ 6 }=\sin \left( 2k \pi+\pi+\frac{ \pi }{ 6 } \right),\sin \left( 2k \pi+2\pi-\frac{ \pi }{ 6 } \right) \]
\[\theta=2k \pi+\frac{ 7\pi }{ 6 },2k \pi+\frac{ 11\pi }{ 6 }\]
where k is a positive integer.