If the expression $\frac{\left[\sin \left(\frac{x}{2} ight)+\cos \left(\frac{x}{2} ight)+i an (x

Question

If the expression $\frac{\left[\sin \left(\frac{x}{2}ight)+\cos \left(\frac{x}{2}ight)+i 	an (x)ight]}{1+2 i \sin \left(\frac{x}{2}ight)}$ is real, then the set of all possible values of $x$ is

Professor PC Thomas Classes · Answer

Let $y=\frac{\left[\sin \left(\frac{x}{2}ight)+\cos \left(\frac{x}{2}ight)+i 	an (x)ight]}{1+2 i \sin \left(\frac{x}{2}ight)} \Rightarrow\left(\frac{\frac{1+i \sqrt{3}}{2}}{\frac{1-i \sqrt{3}}{2}}ight)^{n}=1 \Rightarrow\left(\frac{-\omega^{2}}{-\omega}ight)^{n}=1$\ $=\left[\sin \frac{x}{2}+\cos \frac{x}{2}+i 	an (x)ight]$\ $\frac{\left[1-2 i \sin \frac{x}{2}ight]}{\left[1+2 i \sin \frac{x}{2}ight]\left[1-2 i \sin \frac{x}{2}ight]}$\ $\sin \left(\frac{x}{2}ight)+\cos \left(\frac{x}{2}ight)+2 	an x \sin \frac{x}{2}$\ $=\frac{-i\binom{-	an x+2 \sin ^{2}\left(\frac{x}{2}ight)}{+2 \sin \left(\frac{x}{2}ight) \cos \left(\frac{x}{2}ight)}}{1+4 \sin ^{2} \frac{x}{2}}$ As $y$ is given to be real\ $	herefore-	an x+2 \sin ^{2}\left(\frac{x}{2}ight)+2 \sin \frac{x}{2} \cos \left(\frac{x}{2}ight)=0$\ $\Rightarrow \frac{\sin x}{\cos x}-2 \sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}=0$\ $\Rightarrow \sin x\left[\frac{1}{\cos x}-1ight]-[1-\cos x]=0$\ $\Rightarrow\left(\frac{1-\cos x}{\cos x}ight) \sin x-(1-\cos x)=0$\ $\Rightarrow(1-\cos x)\left[\frac{\sin x}{\cos x}-1ight]=0$\ $\Rightarrow \cos x=1 \quad \Rightarrow \quad x=2 n \pi$\ and $	an x=1 \Rightarrow x=n \pi+\frac{\pi}{4}$\ $	herefore \quad x=2 n \pi, n \pi+\frac{\pi}{4}$.\