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

Of course, $x$ cannot be an integer multiple of $\pi$ plus $\frac{\pi}{2}$. We have here a quadratic equation after a little manipulation, which you, yourself, started.
$$\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$$
After multiplying by $\cos x$, using the identity $1=\sin^2 x+\cos^2 x$, and bringing everything to the left hand side:
$$2\sin^2 x+\sin x-1=0$$
If we replace $y=\sin x$, then you will be able find a familiar expression for $y$:
$$2y^2+y-1 = 0$$
Which gives, $y = \frac{1}{2}$ or $-1$. So, $\sin x$ is equal to either of those two values, which yields $x= 2n\pi+\frac{\pi}{6}$, $x=2n\pi+\frac{5\pi}{6}$ or $x=2n\pi+\frac{\pi}{2}$. However, we ruled out the latter solution initially.
