How would you solve the inequality $\sin x \gt \cos x$? $$\sin x \gt \cos x, \qquad (-2\pi &lt;x &lt;2\pi)$$

I tried an approach saying that $\tan x\gt1$ but apparently the solution, which is $\frac{\pi}{4}&lt;x&lt;\frac{5\pi}{4}$ is not good. 

It's a bit confusing.
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 $\sin x &gt; \cos x$.  It looks like given your comment below your answer that you correctly found when $\sin x &gt; \cos x$ on the interval $x\in [0, 2\pi]$!
for $$0&lt;x&lt;\frac{\pi}{4}, \qquad \sin x&lt;\cos x$$ and  for  $$\frac{\pi}{4}&lt;x&lt;\frac{\pi}{2}, \qquad \sin x&gt;\cos x$$