A) 6
B) 4
C) 3
D) 1
Correct Answer: C
Solution :
\[2{{\sin }^{3}}\alpha -7{{\sin }^{2}}\alpha +7\sin \alpha -2=0\] \[\Rightarrow \]\[2{{\sin }^{2}}\alpha (sin\alpha -1)-5sin\alpha (sin\alpha -1)\] \[+2(sin\alpha -1)=0\] \[\Rightarrow \]\[(sin\alpha -1)(2si{{n}^{2}}\alpha -5sin\alpha +2)=0\] \[\Rightarrow \]\[\sin \alpha -1=0\]or\[2{{\sin }^{2}}\alpha -5\sin \alpha +2=0\] \[\sin \alpha =1\]or\[\sin \alpha =\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm 3}{4}\] \[\alpha =\frac{\pi }{2}\]or\[\sin \alpha =\frac{1}{2},2\]Now,\[\sin \alpha \ne 2\] for, \[\sin \alpha =\frac{1}{2}\] There are three values of a between \[[0,2\pi ]\]
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