Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x&lt;2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}&lt;2$$
$$ 4t^2-2t&gt;0$$
$$ 2t(2t-1)&gt;0$$
$$ t(2t-1)&gt;0$$
$$ (t&gt;0 \wedge t&gt;\dfrac{1}{2}) \vee (t&lt;0 \wedge t&lt;\dfrac{1}{2})$$
From this, I  can only find $x&lt;2\pi+2k\pi$, and, $x&lt;2k\pi$, these are good (I think), but I should find another two solutions.
You have used the substitution:
$$
t=\tan \left(\frac{x}{2}\right)
$$ and the solution of the inequality is $ t&lt;0 \lor t&gt;\dfrac{1}{2}$

so you have:
$$
\tan \left(\frac{x}{2}\right)&lt;0 \qquad \lor \qquad \tan \left(\frac{x}{2}\right)&gt; \dfrac{1}{2}
$$

the solution of the first inequality is:

$$
-\dfrac{\pi}{2}+k\pi&lt;\dfrac{x}{2}&lt;0+k\pi \iff (2k-1)\pi&lt;x&lt;2k\pi
$$

and the solution of the second inequality is:

$$
\tan^{-1}\left(\dfrac{1}{2}\right)+k\pi&lt;\dfrac{x}{2}&lt;\dfrac{\pi}{2}+k\pi \iff 2\tan^{-1}\left(\dfrac{1}{2}\right)+2k\pi&lt;x&lt;\pi+2k\pi 
$$
The answer is the following set $X$.
$$
X=\{t|2n\pi+2\tan^{-1}(\frac{1}{2})&lt;t&lt;2n\pi+2\pi  {\rm ~for~some~} n \in \mathbb{Z}\}
$$