How would you solve the inequality $\sin x \gt \cos x$? $$\sin x \gt \cos x, \qquad (-2\pi <x <2\pi)$$
I tried an approach saying that $\tan x\gt1$ but apparently the solution, which is $\frac{\pi}{4}<x<\frac{5\pi}{4}$ is not good. 
It's a bit confusing. 
 A: First look at where $\sin x = \cos x$ (the unit circle will be of help). 
Here is the unit circle available at Wikipedia. The ordered pairs along the circumference correspond to $\;(\cos \theta, \sin\theta)$

There are two such solutions $x_1, x_2 \in [0, 2\pi]$: $\;\pi/4, 5\pi/4$. $\sin (\pi/4) = \cos(\pi/4) = 1/2,\;$ and $\;\sin(5\pi/4) = \cos (5\pi/4) = -1/2$.
That will give you intervals to check for when $\sin x > \cos x$.  It looks like given your comment below your answer that you correctly found when $\sin x > \cos x$ on the interval $x\in [0, 2\pi]$!
A: Draw (or plot) a picture and take the $2\pi$ periodicity into account.

A: First split the problem in 4 sub-problems regarding each interval.
Reduce every angle to a respective angle in the first quadrant. Then find the angle where both values have same value and use the fact that sine is increasing, while cosine is decreasing function in the first quadrant.
Another way is to shift one angle by 90, and use one well known trigonometric identity:
$$\sin(x)>\cos(x) = \sin(90 -x)$$
$$\sin(x) - \sin(90 - x) >0$$
$$2\sin(x - 45)\cos(45) > 0$$
Now just find when $\sin(x-45) > 0$
Your idea is good, but you don't applied it well, since when you divide by a negative number the sign changes.
A: Over the interval $−2\pi < x < 2\pi$, $\cos x$ as well as $\sin x$ can have negative values. $\tan x>1$ would be good approach for finding $|\sin x|>|\cos x|$.
A: This is a version of the trigonometric circle.
At each point on the circle, the $\;y$-coordinate denotes the value of sine, and the $\;x$- coordinate the value of cosine, and since you've edited your question and you only want the values in the circle and its negative for which the inequality is true, the diagram almost-almost  gives you the complete answer...
Algebraically:
$$\text{For}\;\;x\ge 0:\;\;\;\sin x>\cos x\iff\begin{cases}\frac\pi4<x<\frac{3\pi}4&,\;\;\text{or}\\{}\\
\pi<x<\frac{5\pi}4&,\;\;\text{or}\\{}\\
\frac{7\pi}4<x<2\pi\end{cases}$$
Now, what to do with $\;x<0\;$ ? Well, that's for you now to figure out using that
$$x<0\implies -x>0\;,\;\;\sin(-x)=-\sin x\;,\;\;\cos(-x)=\cos x\ldots$$
A: for $$0<x<\frac{\pi}{4}, \qquad \sin x<\cos x$$ and  for  $$\frac{\pi}{4}<x<\frac{\pi}{2}, \qquad \sin x>\cos x$$ 
