A) \[\frac{1}{2}[n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(3/4)]\]
B) \[n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(3/4)\]
C) \[\frac{n\pi }{2}+{{(-1)}^{n}}{{\sin }^{-1}}(3/4)\]
D) None of these
Correct Answer: A
Solution :
\[3(\sin \theta -\cos \theta )=4\sin \theta \cos \theta \] Þ \[3(\sin \theta -\cos \theta )=2\sin 2\theta \] Squaring both sides, we get \[9(1-S)=4{{S}^{2}},\] where \[S=\sin 2\theta \] or \[4{{S}^{2}}+9S-9=0\]. \[\therefore \] \[\,(S+3)\,(4S-3)=0\] or \[S=\frac{3}{4}\] as \[S\ne -3\] or \[\sin 2\theta =\frac{3}{4}=\sin \alpha \] \[\therefore \] \[2\theta =n\pi +{{(-1)}^{n}}\alpha \] or \[\theta =\frac{1}{2}\,\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\left( \frac{3}{4} \right) \right]\].
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