Let sinx= t 2t2 +3t – 2 = 2t2 +4t -t -2 = 2t(t+2) -1(t+2)= (2t -1)(t+2) Now given (2t-1)(t+2)>0 => t>1/2 or t<-2 => sinx> ½ or sinx <-2 (not possible) So we take first one only x – ( pi/6, 5pi/6) ….….…..(1) For second equation x2 – x-2 <0 = (x-2)(x+1)<0 => -1<x<2 ................. (2) Combining 1 and 2 x- (pi/6, 2) In you have any doubts please reply on this thread.