Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$ Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
$$\tan x + \sec x = 2\cos x$$
I tried changing it all to sin and cos
$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$
then I made it to one fraction
$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$
Then I don't know where to go from there. Please help!

Multiply across by $\cos x$ to get $\sin x + 1 = 2 \cos^2 x= 2 - 2 \sin^2 x$,
or $2 \sin^2 x +\sin x -1 = 0$.
Since the solutions of $2y^2+y-1 = 0$ are $\{-1, {1 \over 2} \}$, the possible solutions to the original equations are $\sin x = -1$ and $\sin x = {1 \over 2} $. However, if $\sin x = -1$, then $\cos x = 0$ and the equation is undefined, hence we ignore this possibility.
The remaining solutions are $\arcsin ( { 1\over 2}) + 2\pi n= { \pi \over 6} + 2 \pi n$.
