The equation $\frac{\mathrm{bcos}\mathrm{x}}{2\cos 2\mathrm{x}-1}=\frac{\mathrm{b}+\sin \mathrm{x}}{\left({\cos }^2\mathrm{x}-3{\sin }^2\mathrm{x}\right)\tan \mathrm{x}}$possess a solution if

The conditions for the existence of a solution are
$$\begin{gathered} 2\cos 2x-1\ne 0\Rightarrow \cos 2x\ne \frac{1}{2}\Rightarrow 2x\ne \frac{\pi }{6} \\ \text{ 2. }\tan x\ne 0\Rightarrow x\ne n\text{ π} \\ \text{ 3. }{\cos }^{2}x-3{\sin }^{2}x\ne 0\Rightarrow {\tan }^{2}x\ne \frac{1}{3}\Rightarrow x\ne \pm \frac{\pi }{6} \\ \text{ As }2\cos x-1=2\left({\cos }^{2}x-{\sin }^{2}x\right)-1 \\ =2\left({\cos }^{2}x-{\sin }^{2}x\right)-\left({\cos }^{2}x+{\sin }^{2}x\right) \\ ={\cos }^{2}x-3{\sin }^{2}x \end{gathered}$$
So the given equation can be written as $\mathrm{bsin}\mathrm{x}=\mathrm{b}+\sin \mathrm{x}$
$\Rightarrow \sin \mathrm{x}=\frac{\mathrm{b}}{\mathrm{b}-1}$
Which is possible if $-1\le \frac{\mathrm{b}}{\mathrm{b}-1}\le 1$
$\begin{array}{l}\Rightarrow \frac{2\mathrm{b}-1}{\mathrm{b}-1}\ge 0\text{ and }\frac{1}{\mathrm{b}-1}\le 0\\ \Rightarrow 2\mathrm{b}-1\le 0\Rightarrow \mathrm{b}\le \frac{1}{2}\end{array}$
But for $\mathrm{b}=\frac{1}{2},\sin \mathrm{x}=-1$which is not possible as $\mathrm{x}\ne -\frac{\mathrm{\pi }}{2}$
Hence equation has a solution $\text{ if }\mathrm{b}<\frac{1}{2}$