Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$ 
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
  $$\tan x+\sec x=2\cos x$$

$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$ 

This is more a comment than an answer, but it is an important comment, since it affects the list of solutions. We seem to have solutions that come from $\sin x=\frac{1}{2}$, and solutions that come from $\sin x=-1$.
No problem with the $\sin x=\frac{1}{2}$ stuff. For the record, this gives the solutions $x=\frac{\pi}{6}+2k\pi$ and $x=\frac{5\pi}{6}+2k\pi$, where $k$ ranges over the integers, positive, negative, and $0$.
However, there is an issue with $\sin x=-1$. For then $\cos x=0$, and neither $\tan x$ nor $\sec x$ is defined. When we multiply through by $\cos x$, we must remember that if $\cos x=0$ we are multiplying both sides by $0$, so we are not getting an equivalent equation. Thus the "solutions" $-\frac{\pi}{2}+2k\pi$ must be discarded.
Remark: If we look at the function $\frac{\sin x+1}{\cos x}$, it turns out that it has a removable singularity at $x=-\frac{\pi}{2}$, and can be made continuous there by defining it to be $0$. In that sense, $x=-\frac{\pi}{2}$ is a solution. However, I doubt it would be accepted as a solution in the context the OP is in.
