Question

The range of $b$ for which the equation $\frac{b\cos x}{2\cos 2x-1}=\frac{b+\sin x}{({\cos }^{2}x-3{\sin }^{2}x)\tan x}$ has real solution(s).

• A. $b\in (-\infty ,\frac{1}{2})-\{-1,0,\frac{1}{3}\}$
• B. $b\in (-\infty ,\frac{1}{2}]\cup (1,\infty )$
• C. $b\in (-\infty ,\frac{1}{2}]-\{-1,0,\frac{1}{3}\}$
• D. $b\in (-\infty ,1)-\{-1,0,\frac{1}{3}\}$

Answer 1

The correct option is A $b\in (-\infty ,\frac{1}{2})-\{-1,0,\frac{1}{3}\}$

$\frac{b\cos x}{2\cos 2x-1}=\frac{b+\sin x}{({\cos }^{2}x-3{\sin }^{2}x)\tan x}$

$2\cos 2x-1\ne 0\Rightarrow x\ne n\pi \pm \frac{\pi }{6}$
$\tan x\ne 0\Rightarrow x\ne n\pi$
${\cos }^{2}x-3{\sin }^{2}x\ne 0\Rightarrow x\ne n\pi \pm \frac{\pi }{6}$

Also,
$2\cos 2x-1=2({\cos }^{2}x-{\sin }^{2}x)-({\cos }^{2}x+{\sin }^{2}x)$
$={\cos }^{2}x-3{\sin }^{2}x$

Hence, the given equation reduces to
$b\sin x=b+\sin x$
$\Rightarrow \sin x=\frac{b}{b-1}$

Now, $-1\le \sin x\le 1$
$\Rightarrow -1\le \frac{b}{b-1}\le 1$
$\Rightarrow \frac{b}{b-1}+1\ge 0$ and $\frac{b}{b-1}-1\le 0$
​​​​​​​$\Rightarrow \frac{2b-1}{b-1}\ge 0$ and ​​​​​​​$\frac{1}{b-1}\le 0$
​​​​​​​$\Rightarrow b\in (-\infty ,\frac{1}{2}]\cup (1,\infty )$ and $b\in (-\infty ,1)$

$\because \sin x\ne 0,\pm \frac{1}{2}$
​​​​​​​$\Rightarrow b\ne 0,-1,-\frac{1}{3}$

When $\sin x=\pm 1$ then $\tan x$ is not defined.
$\therefore b\ne \frac{1}{2}$

​​​​​​​So, ​​​​​​​$b\in (-\infty ,\frac{1}{2})-\{-1,0,\frac{1}{3}\}$