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:$-\infty\lt x\lt\infty$ 

You correctly applied the definitions of $\tan$ and $\sec$ to go from the equation in the problem statement to
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x.$$
However, you combined the two fractions incorrectly; in general
$$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$$
For example, you know that 
$$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$
and that $$\frac{1}{4}+\frac{1}{4}\neq\frac{1}{4}.$$

The correct form of the equation should be
$$2\cos^2x-\sin x-1=0.$$
You will want to use the fact that for any $x$,
$$\sin^2(x)+\cos^2(x)=1.$$

 Thus, any occurrence of $\cos^2(x)$ can be replaced with $(1-\sin^2(x))$. $$2(1-\sin^2(x))-\sin(x)-1=0$$ $$[2\cdot1-2\cdot\sin^2(x)]-\sin(x)-1=0$$ $$-2\cdot\sin^2(x)+(-1)\sin(x)-1+2=0$$ $$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$

Then, you will need to use the quadratic formula to solve for $\sin(x)$. It may help to write $y=\sin(x)$ temporarily, to prevent confusion. Keep in mind that there may be two different possible values of $\sin(x)$ that result. 

 $$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$ $$(-2)y^2+(-1)y+(1)=0$$ $$y=\frac{-(-1)\pm\sqrt{(-1)^2-4(-2)(1)}}{2(-2)}=\frac{1\pm\sqrt{1+8}}{-4}=\frac{1\pm 3}{-4}=\begin{cases}\frac{1+3}{-4}=\frac{4}{-4}=\fbox{$-1$} &\text{ or }\\\\ & \\ \frac{1-3}{-4}=\frac{-2}{-4}=\fbox{$\frac{1}{2}$}.\end{cases}$$

Lastly, you need to find those values of $x$ that produce that value (or those values) of $\sin(x)$. Keep in mind that the function $\sin(x)$ is periodic; here is part of its graph to illustrate:

(and it continues similarly out to infinity in each direction) so that there are always infinitely many values of $x$ that will produce a given value of $\sin(x)$. 
For any real number $x$, the values of
$$\sin(x+2\pi k)$$
are all identical for any integer $k$,. There are usually even more values you can plug into $\sin$ that will give the same number (look at the graph for an idea).
