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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
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regards
Ramesh
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