Question

The value of ${}^{\prime }{b}^{\prime }$ such that the equation $\frac{b\cos x}{2\cos 2x-1}=\frac{b+\sin x}{({\cos }^{2}x-3{\sin }^{2}x)\tan x}$ possess solution, then prove that b belongs to the set $(-\infty ,\frac{1}{2})$.

Answer 1

For the domain of definition of the given equation, we have
(i) $2\cos 2x-1\ne 0\Rightarrow x\ne n\pi \pm \frac{\pi }{6}$
(ii) $\tan x\ne 0\Rightarrow x\ne \pm \frac{n\pi }{2}$ $[$ For odd multiples of $\frac{\pi }{2},\tan x$ is not defined $]$
(iii) ${\cos }^{2}x-3{\sin }^{2}x\ne 0\Rightarrow x\ne n\pi \pm \frac{\pi }{6}$
Also, $2\cos 2x-1=2({\cos }^{2}x-{\sin }^{2}x)-({\cos }^{2}x+{\sin }^{2}x)={\cos }^{2}x-3{\sin }^{2}x$
Now, the given equation reduces to
$b\sin x=b+\sin x\Rightarrow \sin x=\frac{b}{b-1}$
$\because -1\le \sin x\le 1$
$\therefore -1\le \frac{b}{b-1}\le 1$
$\Rightarrow \frac{b}{b-1}+1\le 0$ and $\frac{b}{b-1}-1\le 0$
$\Rightarrow \frac{2b-1}{b-1}\ge 0$ and $\frac{1}{b-1}\le 0$
$\Rightarrow b\le \frac{1}{2}\Rightarrow b>1$ and $b<1\Rightarrow b\le \frac{1}{2}$
when $b=\frac{1}{2}\Rightarrow \sin x=1$, which is not possible
$\therefore b<\frac{1}{2}$