3 sin^x + 4 sinx + 1 = 0

[latex]3 sin^2x + 4 sinx + 1 = 0 \\ D=16-4*3*1=16-12=4 \\ sinx_1= \frac{-4+2}{6} = -\frac{2}{6} =- \frac{1}{3} ~~~~~~~~~~~~~~~~~~sinx_2= \frac{-4-2}{6}=- \frac{6}{6} =-1 \\ sinx=- \frac{1}{3} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~sinx=-1 \\ \\ x=(-1)^{k+1} arcsin\frac{1}{3} + \pi k~~~~~~~~~~~x=- \frac{ \pi }{2} +2 \pi k \\ k\in Z[/latex]