A) 0
B) 2
C) 3
D) infinite
Correct Answer: D
Solution :
\[a\cos 2x+b{{\sin }^{2}}x+c=0\] \[\Rightarrow \] \[a(1-2{{\sin }^{2}}x)b{{\sin }^{2}}x+c=0\] \[ie\], \[(b-2a){{\sin }^{2}}x+(a+c)=0\] It is an identity, if\[b-2a=0,\,\,a+c=0\], so \[\frac{a}{1}=\frac{b}{2}=\frac{c}{-1}\] Thus, number of triplet\[(a,\,\,b,\,\,c)\]are infinite.You need to login to perform this action.
You will be redirected in
3 sec