Trigonometrical Equations 2 Question 3
3. Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta$ $\tan \frac{\theta}{2}+\cot \frac{\theta}{2} \cos \varphi-1, \tan (2 \pi-\theta)>0$ and $-1<\sin \theta<-\frac{\sqrt{3}}{2}$. Then, $\varphi$ cannot satisfy
(2012)
(a) $0<\varphi<\frac{\pi}{2}$
(b) $\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$
(c) $\frac{4 \pi}{3}<\varphi<\frac{3 \pi}{2}$
(d) $\frac{3 \pi}{2}<\varphi<2 \pi$
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Answer:
Correct Answer: 3. $(a, c, d)$
Solution:
- PLAN It is based on range of $\sin x$, i.e. $[-1,1]$ and the internal for $a<x<b$.
Description of Situation As $\theta, \varphi \in[0,2 \pi]$ and $\tan (2 \pi-\theta)>0,-1<\sin \theta<-\frac{\sqrt{3}}{2}$ $\tan (2 \pi-\theta)>0$ $-\tan \theta>0$
$\Rightarrow \theta \in I I$ or IV quadrant.
Also, $\quad-1<\sin \theta<-\frac{\sqrt{3}}{2}$
$\Rightarrow \quad \frac{4 \pi}{3}<\theta<\frac{5 \pi}{3}$ but $\theta \in$ II or IV quadrant
$\Rightarrow \quad \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3}$
Here, $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \tan \frac{\theta}{2}+\cot \frac{\theta}{2} \cos \varphi-1$
$\Rightarrow 2 \cos \theta-2 \cos \theta \sin \varphi=\sin ^{2} \theta \frac{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}} \cos \varphi-1$
$\Rightarrow 2 \cos \theta-2 \cos \theta \sin \varphi=2 \sin ^{2} \theta \frac{1}{\sin \theta} \cos \varphi-1$
$\Rightarrow \quad 2 \cos \theta+1=2 \sin \varphi \cos \theta+2 \sin \theta \cos \varphi$
$\Rightarrow \quad 2 \cos \theta+1=2 \sin (\theta+\varphi)$
From Eq. (i), $\frac{3 \pi}{2}<\theta<\frac{5 \pi}{3}$
$\Rightarrow \quad 2 \cos \theta+1 \in(1,2)$
$\therefore \quad 1<2 \sin (\theta+\varphi)<2$
$\Rightarrow \quad \frac{1}{2}<\sin (\theta+\varphi)<1$
$\Rightarrow \quad \frac{\pi}{6}<\theta+\varphi<\frac{5 \pi}{6}$
or $\quad \frac{13 \pi}{6}<\theta+\varphi<\frac{17 \pi}{6}$
$\therefore \quad \frac{\pi}{6}-\theta<\varphi<\frac{5 \pi}{6}-\theta$
or $\quad \frac{13 \pi}{6}-\theta<\varphi<\frac{17 \pi}{6}-\theta$
$\Rightarrow \varphi \in-\frac{3 \pi}{2},-\frac{2 \pi}{3}$ or $\frac{2 \pi}{3}, \frac{7 \pi}{6}$, as $\theta \in \frac{3 \pi}{2}, \frac{5 \pi}{3}$
