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.
 A: 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.
A: Hint:  if you put the two fractions over a common denominator you get a nice cancellation.
A: You have a fraction and its reciprocal adding to $2$: $\lambda + 1/\lambda=2$. Only solution is $\lambda=1$, as @J.M. has noted. So $\cos x = 1+\sin x$. Draw the graphs of the two sides of the equation and see that there are five intersections in the $x$-interval $[-2\pi,2\pi]$, but the two at $x=-\pi/2$ and $x=3\pi/2$ are no good, ’cause there you get $0/0$ occurring in the original problem.
A: $$ \begin {align*}
&  2=\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x}\\
& =\frac{\cos^2 x+(1+\sin x)^2}{\cos x(1+\sin x)}\\
& =\frac{\cos^2 x+\sin^2x+1+2\sin x}{\cos x(1+\sin x)}\\
& =\frac{2+2 \sin x}{\cos x(1+\sin x)} \, \\
& =\frac{2( 1+\sin x) }{\cos x(1+\sin x)}\;\\
& =\frac{2}{\cos x}\;
\end{align*}$$
$$ \rightarrow \cos x = 1, x = 2 k \pi  $$  
A: After simplification, with condition $1 + \sin x \ne 0$ (or $x \ne 3\pi/2$),  the given equation becomes:
$$\frac{\cos x}{(1 + \sin x)} + \frac{(1+ \sin x)}{\cos x} = \frac{2}{\cos x} = 2\;,$$
which implies that
$$\cos x = \frac22 = 1\;.$$
Calculators and the Trig Unit Circle give as answers:
$x = 0$ (or $0$ deg.) and  $x = 2\pi$ (or $360$ deg.)
Let's check: when $x = 0$ or $x = 2\pi$, we have $\cos x = 1$ and $\sin x = 0$, then:
$1/1 + 1/1 = 2$ (True).
A: As $\cos x\cdot\cos x=(1-\sin x)(1+\sin x)$
$$\dfrac{\cos x}{1+\sin x}=\dfrac{1-\sin x}{\cos x}$$
So, the given relation reduces to $$\dfrac2{\cos x}=2\iff\cos x=1$$
