A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: D
Solution :
Given, \[{{\sin }^{2}}\left( \frac{\pi }{8}+\alpha \right)-{{\sin }^{2}}\left( \frac{\pi }{8}-\alpha \right)=\frac{1}{2\sqrt{2}}\] \[\Rightarrow \]\[\sin \left( \frac{\pi }{8}+\alpha +\frac{\pi }{8}-\alpha \right)\sin \left( \frac{\pi }{8}+\alpha -\frac{\pi }{8}+\alpha \right)=\frac{1}{2\sqrt{2}}\] \[[\because \,\,{{\sin }^{2}}x-{{\sin }^{2}}y=\sin (x+y)\sin (x-y)]\] \[\Rightarrow \] \[\sin \left( \frac{2\pi }{8} \right)\sin (2\alpha )=\frac{1}{2\sqrt{2}}\] \[\Rightarrow \] \[\sin \frac{\pi }{4}\sin 2\alpha =\frac{1}{2\sqrt{2}}\] \[\Rightarrow \] \[\frac{1}{\,\sqrt{2}}\sin 2\alpha =\frac{1}{2\sqrt{2}}\] \[\Rightarrow \] \[\sin 2\alpha =\frac{1}{2}\] \[\Rightarrow \] \[\sin 2\alpha =\sin \frac{\pi }{6}\] \[\Rightarrow \] \[2\alpha =n\pi +{{(-1)}^{n}}.\frac{\pi }{6}\] \[\Rightarrow \] \[\alpha =\frac{n\pi }{2}+{{(-1)}^{n}}\frac{\pi }{12},\] where \[n\in I\] when \[n=0,\] then \[\alpha =\frac{\pi }{12}\in (-\pi ,\pi )\] when \[n=1,\] then \[\alpha =\frac{\pi }{2}-\frac{\pi }{12}\] \[=\frac{5\pi }{12}\in (-\pi ,\pi )\] when \[n=2,\] then \[\alpha =\pi +\frac{\pi }{12}\] \[=\frac{13\pi }{6}\in /[-\pi ,\pi ]\] When \[n=-1,\] then \[\alpha =-\frac{\pi }{2}-\frac{\pi }{12}\] \[=\frac{-6-1}{12}\pi =\frac{-7\pi }{12}\in [-\pi ,\pi ]\] When \[n=-2,\] then \[\alpha =-\pi +\frac{\pi }{12}=\frac{-11\pi }{12}\in [-\pi ,\pi ]\] Thus, values satisfied in \[[-\pi ,\,\,\pi ]\] are \[\frac{\pi }{12},\frac{5\pi }{12},\frac{-7\pi }{12},\frac{-11\pi }{12}\] Hence, the number of values of \[\alpha \]are 4.
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