A) \[\theta =n\pi \pm \frac{\pi }{6},n\in I\]
B) \[n\pi +\frac{\pi }{6},n\in I\]
C) \[2n\pi +\frac{\pi }{6},n\in I\]
D) \[2n\pi +\frac{11\pi }{6},n\in I\]
Correct Answer: D
Solution :
\[2{{\cot }^{2}}\theta +2\sqrt{3}\cot \theta +4\cos ec\theta +8=0\] \[\Rightarrow \,\,{{\cot }^{2}}\theta +\cos e{{c}^{2}}\theta -1+2\sqrt{3}\cot \theta +4\cos ec\theta +8=0\] \[\left( \because \,\,\,{{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta -1 \right)\] \[\Rightarrow \,\,\left( {{\cot }^{2}}\theta +2\sqrt{3}\cot \theta +3 \right)+\left( \cos e{{c}^{2}}\theta +4\cos ec\theta +4 \right)=0\]\[\Rightarrow \,\,{{\left( \cot \theta +\sqrt{3} \right)}^{2}}+{{\left( \cos ec\theta +2 \right)}^{2}}=0\] \[\Rightarrow \,\,\,\cot \theta =-\sqrt{3}\] and \[\cos ec\theta =-2\] Principal value of \[\theta \] satisfying both the equation is \[\theta =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}\] \[\therefore \] General solution is \[\theta =2n\pi +\frac{11\pi }{6},n\in I\]
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