Question

Find all solutions of the equation in the interval [0,2pi). 2sin^2x+3sinx+1=0

Answer 1

\[(2\sin x+1)(\sin x+1)=0\]

Answer 2

maybe i dunno

Answer 3

nevermind completely.

Answer 4

read it way wrong

Answer 5

\[\sin x=\frac{-1}{2} sinx=-1\]

Answer 6

\[x=\frac{7\pi}{6}, \frac{11\pi}{6}, \frac{3\pi}{2}\]