The number of solutions of the equation  $\mathrm{tanx}+\mathrm{secx}=2\mathrm{cosx}$ lying in the interval $[0,2\mathrm{\pi }]$ is

Here, $$\begin{gathered} \tan x+\sec x=2\cos x \left(x\ne \frac{\mathrm{\pi }}{2}\right) \\ 1+\sin x=2 {\cos }^{2}x \\ \Rightarrow 2{\sin }^2\mathrm{x}+\sin \mathrm{x}-1=0 \\ \Rightarrow \sin \mathrm{x}=\frac{1}{2},-1.\mathrm{But} \sin \mathrm{x}=-1\Rightarrow \mathrm{x}=\frac{3\mathrm{\pi }}{2} \end{gathered}$$
Which does not satisfy the equation.