Complex Numbers Ques 39
If the expression $\frac{[\sin \frac{x}{2}+\cos \frac{x}{2}-i \tan (x)]}{[1+2 i \sin \frac{x}{2}]}$
is real, then the set of all possible values of $x$ is… .
$(1987,2 M)$
Show Answer
Answer:
Correct Answer: 39.$x=2 n \pi+2 \alpha, \alpha$ $=\tan ^{-1} k, \text { where } k \in(1,2) \text { or } x=2 n \pi$
Solution:
- $\frac{(\sin \frac{x}{2}+\cos \frac{x}{2})-i \tan x}{1+2 i \sin \frac{x}{2}} \in R$
$ =\frac{(\sin \frac{x}{2}+\cos \frac{x}{2}-i \tan x) \quad (1-2 i \sin \frac{x}{2})}{1+4 \sin ^{2} \frac{x}{2}} $
Since, it is real, so imaginary part will be zero.
$ \begin{aligned} & \therefore \quad-2 \sin \frac{x}{2} (\sin \frac{x}{2}+\cos \frac{x}{2})-\tan x=0 \\ & \Rightarrow \quad 2 \sin \frac{x}{2} (\sin \frac{x}{2}+\cos \frac{x}{2}) \cos x+2 \sin \frac{x}{2} \cos \frac{x}{2}=0 \\ & \Rightarrow \quad \sin \frac{x}{2} [(\sin \frac{x}{2}+\cos \frac{x}{2}) (\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2})+\cos \frac{x}{2}]=0 \\ & \therefore \quad \sin \frac{x}{2}=0 \\ & \Rightarrow \quad (\sin \frac{x}{2}+\cos \frac{x}{2} )(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2})+\cos \frac{x}{2}=0 \end{aligned} $
On dividing by $\cos ^{3} \frac{x}{2}$, we get
$ \begin{aligned} & (\tan \frac{x}{2}+1) \quad (1-\tan ^{2} \frac{x}{2})+(1+\tan ^{2} \frac{x}{2})=0 \\ & \Rightarrow \quad \tan ^{3} \frac{x}{2}-\tan \frac{x}{2}-2=0 \\ & \text { Let } \quad \tan \frac{x}{2}=t \\ & \text { and } \quad f(t)=t^{3}-t-2 \\ & \text { Then, } \quad f(1)=-2<0 \\ & \text { and } \quad f(2)=4>0 \end{aligned} $
Thus, $f(t)$ changes sign from negative to positive in the interval $(1,2)$.
$\therefore$ Let $t=k$ be the root for which
$ \begin{array}{lc} & f(k)=0 \text { and } \quad k \in(1,2) \\ \therefore & t=k \quad \text { or } \quad \tan \frac{x}{2}=k=\tan \alpha \\ \Rightarrow & x / 2=n \pi+\alpha \\ \Rightarrow & x=2 n \pi+2 \alpha, \alpha=\tan ^{-1} k, \text { where } k \in(1,2) \\ & \text { or } x=2 n \pi \end{array} $
BETA
