A) 0
B) \[-\frac{\pi }{3}\]
C) \[-\pi \]
D) All of these
Correct Answer: B
Solution :
\[\sin \alpha +\int_{\alpha }^{2\alpha }{\cos 2xdx=0}\] \[\Rightarrow \]\[\sin \alpha \frac{1}{2}[\sin 2x]_{\alpha }^{2\alpha }=0\] \[\Rightarrow \]\[\sin \alpha +\frac{1}{2}(sin4\alpha -\sin 2\alpha )=0\] \[\Rightarrow \]\[\sin \alpha +\sin \alpha \cos 3\alpha =0\] \[\Rightarrow \]\[\sin \alpha (1+cos3\alpha )=0\] \[\Rightarrow \]\[\sin \alpha =0\]or \[1+\cos 3\alpha =0\] If \[\sin \alpha =0\] \[\Rightarrow \]\[\alpha =0,\,\pi -\pi \] If \[1+\cos 3\alpha =0\] \[\Rightarrow \] \[\alpha =\frac{-\pi }{3}\] But, only \[-\frac{\pi }{3}\in (-\pi ,0)\] Hence, value of \[\alpha \] is \[-\frac{\pi }{3}.\]
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