Question

4sin^2 x - 8 sin x +3=0

Answer 1

$4sin^2x-8sin x+3=0;\ sin x=t;\ -1leq tleq 1;\ 4t^2-8t+3=0;\ D=8^2-4cdot4cdot3=64-48=16=(pm4)^2;\ t_1=frac{8-4}{8}=frac12in-1leq tleq1;\ t_2=frac{8+4}{8}=frac{3}{2}notin-1leq tleq 1;\ sin x=frac12;\ x=(-1)fracpi6+pi n, nin Z$