Solve $\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2$ I am fairly good at solving trig equations yet this one equation has me stumped. I've been trying very hard but was unable to solve it. Can anyone help please? Thank you.
$$\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2$$
solve for $x$ in the range of $[-2\pi, 2\pi]$
I do know we have to do a difference of squares, yet after that, I don't know what to do and I get lost.
Thank you.

HINT: $$\begin{align*}
\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x}&=\frac{\cos^2 x+(1+\sin x)^2}{\cos x(1+\sin x)}\\
&=\frac{\cos^2 x+\sin^2x+1+2\sin x}{\cos x(1+\sin x)}\;;
\end{align*}$$
now use a familiar trig identity and find something to cancel.
