A) 0
B) 1
C) 3
D) \[\infty \]
Correct Answer: D
Solution :
Given, \[{{a}_{1}}+{{q}_{2}}\cos 2x+{{a}_{3}}{{\sin }^{2}}x=0,\]for all x \[\Rightarrow \]\[{{a}_{1}}+{{a}_{2}}\cos 2x+{{a}_{3}}\left( \frac{1-\cos 2x}{2} \right)=0,\]for all x \[\Rightarrow \]\[\left( {{a}_{1}}+\frac{{{a}_{3}}}{2} \right)+\left( {{a}_{2}}-\frac{{{a}_{3}}}{2} \right)\cos 2x=0,\]for all x \[\Rightarrow \]\[{{a}_{1}}+\frac{1}{2}{{a}_{3}}=0\]and \[{{a}_{2}}-\frac{{{a}_{3}}}{2}=0\] \[\Rightarrow \]\[{{a}_{1}}=-\frac{k}{2},{{a}_{2}}=\frac{k}{2},{{a}_{3}}=k,\] where k is any real number. Thus, the number of triplets is infinite.
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