A) \[0\]
B) \[1\]
C) \[3\]
D) infinitely many
Correct Answer: D
Solution :
We have, \[a+b\cos 2x+c{{\sin }^{2}}x=0\]for all\[x\] \[\Rightarrow \] \[a+b(1-2{{\sin }^{2}}x)+c{{\sin }^{2}}x=0\] \[\Rightarrow \] \[a+b+(c-2b){{\sin }^{2}}x=0\]for all\[x\] \[\Rightarrow \] \[a+b=0\]and\[c-2b=0\] Thus, the triplets are\[(-b,\,\,b,\,\,2b)\], where\[b\in R\]. Hence, there are infinitely many triplets.
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