Question

find the number of all possible triplets(a1,a2,a3) such thata1+a2sin2x+a3sin^2x=0.

Answer 1

is this ur question?\[a_1+a_2 \sin 2x+a_3 \ \sin^2 x=0\]

Answer 2

\[\frac{ -2a1 - a3 }{ 2 } = a2\sin2x - \frac{ a3 }{ 2 }\cos2x \]

Answer 3

Clearly only visible condition on a1 , a2, a3 is that:

\[-1 \le \frac{ 2a1 + a3 }{ \sqrt{4(a2)^2 + (a3)^2} }\]

Answer 4

\[\le1\]

Answer 5

MAybe I'm missing something. As this still gives infinite cases.

Answer 6

i think its supposed to be \[a_1+a_2 \cos 2x+a_3 \ \sin^2 x=0\]more likely

Answer 7

Okay.
That is probable.

Answer 8

just a thought :)