<s>[INST] Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I  can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), but I should find another two solutions.
 [/INST] First, we solve $\sin x + 2 \cos x = 2$
Let $\theta$ be the angle corresponding to the point $(x,y) = (2,1)$ with amplitude $r = \sqrt 5.$ Then
$\cos \theta = \dfrac{2}{\sqrt 5}$ and
$\sin \theta = \dfrac{1}{\sqrt 5}$
Then
\begin{align}
   \dfrac{2}{\sqrt 5}\cos x + \dfrac{1}{\sqrt 5} \sin x &= \dfrac{2}{\sqrt 5}\\
   \cos x \cos \theta + \sin x \sin \theta &= \dfrac{2}{\sqrt 5}\\
   \cos(x - \theta) &= \dfrac{2}{\sqrt 5}\\
   \cos(x - \theta) &= \cos \theta\\
   x - \theta &= 2n\pi \pm \theta\\
   x &\in \{2n\pi + 2\theta : n \in \mathbb Z\} \cup
          \{2n\pi : n \in \mathbb Z\}
\end{align}
This splits the unit circle, $x \in [0, 2\pi]$, into four pieces:
$\{0\},\; (0, 2\theta),\; \{2\theta\},\; (2\theta, 2\pi)$
Equality occurs at $x = 0$ and $x = 2\theta$.
$\sin x + 2 \cos x < 2$ will occur in the interval
$(0, 2\theta)$ or the interval
$(2\theta, 2\pi).$
Since $\theta$ is in the first quadrant, $\theta \in (0, 2\theta).$
$\sin \theta + 2 \cos \theta = \sqrt 5 > 2$
So the intervals $(2n\pi, 2n\pi + 2\theta)$ are not part of the solution set.
Since $\theta$ is in the first quadrant, $\pi \in (2\theta, 2\pi).$
$\sin \pi + 2 \cos \pi = -2 < 2$
So the intervals $(2n\pi+2\theta, (2n+1)\pi)$ are part of the solution set.
So the solution set is $x \in (2n\pi+2\theta, (2n+1)\pi)\quad \forall n \in \mathbb Z$ where $\theta = \arcsin \dfrac{1}{\sqrt 5}$
</s>