A) \[\frac{7\pi }{12}\]
B) \[\frac{7\pi }{6}\]
C) \[\frac{7\pi }{3}\]
D) \[\frac{7\pi }{2}\]
Correct Answer: A
Solution :
\[\frac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }+\frac{{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta }=12+\frac{8}{{{\sin }^{3}}2\theta }\] \[\Rightarrow \] \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\frac{3}{2}{{\sin }^{3}}2\theta +1\] \[\Rightarrow \] \[1-\frac{3}{4}{{\sin }^{2}}2\theta =\frac{3}{2}{{\sin }^{3}}2\theta +1\] \[\Rightarrow \] \[\sin \,2\theta \,({{\sin }^{2}}2\theta +2\,\sin \,2\theta +1\] \[\Rightarrow \] \[\sin \,2\theta \,=-\frac{1}{2}\,=\,\sin \,\left( -\frac{\pi }{6} \right)\] \[(\because \,\,\sin \,2\theta \ne 0)\] \[\Rightarrow \] \[2\theta =n\pi +{{(-1)}^{n}}\left( -\frac{\pi }{6} \right)\] \[\Rightarrow \] \[\theta =\frac{n\pi }{2}+\frac{\pi }{12}{{(-1)}^{n+1}}\] \[\Rightarrow \] \[\theta =\frac{7\pi }{12},\,\frac{11\pi }{12},\,\frac{19\pi }{12},\,\frac{23\pi }{12}\]
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