The number of all possible triplets (a1, a2,a3) such that a1 + a2 cos2x +a3 sin ² x = 0 for all x is

(a) 0 (b) 1 (c)3 (d) infinity

A + B(1-t²) + Ct²      where t² = sin²x   , i.e., 0 =< t² =< 1 as -1<sin x<1

t² = (2B-C)/(A+B)

0 =< (2B-C)/(A+B) =< 1

i.e., 2B >= C  and A+C>=B

the possible triplets for which the above inequalities satisfy are :

(A,B,C) in order of (0,1,1) (1,1,0) (0,0,0) (2,3,1) and so on

answer is infinity

--

regards

Ramesh