Find all values of $x$ in the interval $[0,2\pi]$ that satisfy $\sin x + \tan x =1+ \sec x$ I've been doing some practice questions for a test, but I'm having issues with a a particular one.
Find all values of $x$ in the interval $[0,2\pi]$ that satisfy $$\sin x + \tan x =1+ \sec x$$
I wasn't sure what to do so I tried simplifying it, and I got it down to $$\sin x = 1$$
For which I would believe the answer is $\pi/2$ but the correct answer I'm told is just $\pi$. Have I messed up somewhere, or is the answer I was given incorrect?
Thanks
 A: TLDR: You forgot to account for the fact that, in dividing by $\cos x + 1$, you assume that it is non-zero.
$$\begin{align}
&\sin x + \frac{\sin x}{\cos x} =1+ \frac{1}{\cos x}\\
&\implies \sin x\cos x + \sin x = \cos x + 1\\
&\implies \sin x(\cos x + 1) = \cos x + 1\\
&\implies \sin x = 1 \quad \lor \quad \cos x  = -1\\
\end{align}$$
We now note that the first equality is solved when $x = \pi (n + \frac{1}{2})$ and the second is solved when $x = \pi(2n+1)$. However, we note that solutions to $\sin x = 1$ yield $+\infty$ on both the RHS and LHS of the original equation; Whether or not you define this as equality is somewhat contextual, as subtracting the RHS from the LHS to get $y =\sin x + \tan x - \sec x - 1$ yields zeroes at both $\frac \pi 2$ as well as at $\pi$. However, accepting only finite equality we can choose to ignore the solutions of the form $x = \pi (n + \frac{1}{2})$. Therefore, we only need to solve the RHS equality in the domain $[0,2\pi]$, with the only answer being $x= \pi$  
Note: There is an interesting relationship between the RHS and LHS of your original equation at half multiplies of $\pi$, i.e. solutions to $\sin x = 1$. If we take the limit of the quotient of the RHS and LHS, i.e. $\lim_{x \to \pi}\frac{\sin x + \tan x}{1 + \sec x}$ we find that the answer is $1$, which means that, although both blow up to infinity at $\frac \pi 2$, they blow up at asymptotically the same rate. 
