A) \[n\pi +\alpha \]
B) \[2n\pi \]
C) \[\frac{n\pi }{2}+\alpha \]
D) None of these
Correct Answer: B
Solution :
Since, \[\frac{\left( \sin \frac{x}{2}+cos\frac{x}{2} \right)-i\tan x}{1+2i\sin \frac{x}{2}}\in R\] \[\Rightarrow \]\[\frac{\left\{ \sin \frac{x}{2}+\cos \frac{x}{2}-i\tan x \right\}\left\{ 1-2i\sin \frac{x}{2} \right\}}{1+4{{\sin }^{2}}\frac{x}{2}}\in R\] It will be real, if imaginary part is zero. \[\therefore \] \[-2\sin \frac{x}{2}\left\{ \sin \frac{x}{2}+\cos \frac{x}{2} \right\}-\tan x=0\] \[\Rightarrow \] \[2\sin \frac{x}{2}\left\{ \sin \frac{x}{2}+\cos \frac{x}{2} \right\}-\frac{\sin x}{\cos x}=0\] \[\Rightarrow \] \[\sin \frac{x}{2}\left[ \left\{ \sin \frac{x}{2}+\cos \frac{x}{2} \right\} \right.\left\{ {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} \right\}\] \[\left. +\cos \frac{x}{2} \right]=0\] \[\therefore \] \[\sin \frac{x}{2}=0\] \[\Rightarrow \] \[x=2n\pi \] ...(i) Or \[\left\{ \sin \frac{x}{2}+\cos \frac{x}{2} \right\}\left\{ {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} \right\}\] \[+\cos \frac{x}{2}=0\] On dividing by \[{{\cos }^{3}}\frac{x}{2}\] \[\left( \tan \frac{x}{2}+1 \right)\left( 1-{{\tan }^{2}}\frac{x}{2} \right)+\left( 1+{{\tan }^{2}}\frac{x}{2} \right)=0\] \[\Rightarrow \] \[{{\tan }^{3}}\frac{x}{2}-\tan \frac{x}{2}-2=0\] Let \[\tan \frac{x}{2}=t,\] then \[f(t)={{t}^{3}}-t-2,\] then\[f(1)=-2<0\]and\[f(2)=4>0\] Thus,\[f(t)\]changes sign from negative to positive in (1, 2). \[\therefore \]Let\[t=k\]be the root for which \[f(k)=0\] and \[k\in (1,2)\] \[\therefore \] \[t=k\]or \[\tan \frac{x}{2}=k=\tan \alpha \] Hence, \[\frac{x}{2}=n\pi +\alpha \] \[\Rightarrow \]\[\left\{ \begin{matrix} x=2n\pi +2\alpha ,\alpha ={{\tan }^{-1}}k \\ or\,x=2n\pi \\ \end{matrix} \right.\]where \[k\in (1,2)\]
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