Sir please clear my doubt in quadratic equations. i face problem in dealing with such questions. one is here:- Find the values of m for which the equation sin^2x-(m-3)sinx+m=0 has real roots. I can assume sinx to be t and reduce the equation to simple for. now t is binded in the interval -1,1. Now how to do sir? please clear! Share with your friends Share 0 Lovina Kansal answered this sin2x-(m-3)sinx+m=0Let sinx=yy2-(m-3)y+m=0For real roots,D=b2-4ac>0So, -m-32-4(1)(m)>0⇒m2+9-6m-4m>0⇒m2-10m+9>0⇒m2-m-9m+9=>0⇒(m-9)(m-1)>0Compute the signs of the factors of (m-9)(m-1) m<1 m=1 1<m<9 m=9 m>9 m-9 - - - 0 + m-1 - 0 + + + (m-9)(m-1) + 0 - 0 + Choosing the ranges that satisfy the required condition:>0m<1 or m>9i.e m∈(-∞,1)∪9,∞ -1 View Full Answer