Find the number of sets of 5-tuples (A1,A2,A3,A4,A5) so that A1 + A2(sinx)+ A3(cosx) + A4(2cosx) + A5(2sinx) = 0 is true for all values of X? A) 0 B) 1 C) 2 D) Infinite

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(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})

a_{1} + a_{2} s i n x + a_{3} c o s x + a_{4} s i n 2 x + a_{5} c o s 2 x = 0

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(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})

a_{1} + a_{2} sin x + a_{3} cos x + a_{4} sin 2 x + a_{5} cos 2 x = 0

x

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

The number of all the possible triplets $(a_{1}, a_{2}, a_{3})$ such that $a_{1} + a_{2} c o s (2 x) + a_{3} s i n^{2} (x) = 0$ for all x is

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} a n d a_{6}

a_{6} > a_{1}, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = p^{3} a n d a_{5} = 39

(a_{3} + a_{4} - 12 p)

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