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

\begin{eqnarray}
  \left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)&=&2\cos x\\
  \sin x + 1 &=& 2 \cos^2 x \\
  \sin x + 1 &=& 2(1-\sin^2 x)\\
\end{eqnarray}
Then $\sin x = -1$ or $\sin x = 1/2$ and the solutions are $-\pi + 2k \pi, \pi/6 + 2 k\pi$  and $5\pi/6 + 2k \pi$.
