A) 3
B) -3
C) 0
D) None
Correct Answer: C
Solution :
Since \[\alpha \] is a common root \[{{\alpha }^{2}}-a\alpha +12=0\] (1) \[{{\alpha }^{2}}-b\alpha +15=0\] (2) \[{{\alpha }^{2}}-(a+b)\alpha +36=0\] \[(1)+(2)-(3)\]\[\Rightarrow \]\[{{\alpha }^{2}}-9=0\Rightarrow \alpha =3\] Given, \[\cos x+\cos 2x+\cos 3x=\alpha =3\] This possible only when \[\cos x=1,\cos 2x=1,os3x=1\] \[\therefore \]\[\sin x=0,\sin 2x=0\And \sin 3x=0\] Hence \[\sin x+\sin 2x+\sin 3x=0\]You need to login to perform this action.
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