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.
 A: You have used the substitution:
$$
t=\tan \left(\frac{x}{2}\right)
$$ and the solution of the inequality is $ t<0 \lor t>\dfrac{1}{2}$
so you have:
$$
\tan \left(\frac{x}{2}\right)<0 \qquad \lor \qquad \tan \left(\frac{x}{2}\right)> \dfrac{1}{2}
$$
the solution of the first inequality is:
$$
-\dfrac{\pi}{2}+k\pi<\dfrac{x}{2}<0+k\pi \iff (2k-1)\pi<x<2k\pi
$$
and the solution of the second inequality is:
$$
\tan^{-1}\left(\dfrac{1}{2}\right)+k\pi<\dfrac{x}{2}<\dfrac{\pi}{2}+k\pi \iff 2\tan^{-1}\left(\dfrac{1}{2}\right)+2k\pi<x<\pi+2k\pi 
$$
A: $$ \sin ( x+ \cos^{-1} \dfrac{1}{\sqrt5})=\dfrac{2}{\sqrt5} ,\; or, \sin ( x+ \tan^{-1} 2)=\frac{2}{\sqrt5}.$$
which you can further. 
A: $\sin x+2\cos x=2(\frac{1}{2}\sin x + \cos x)$
Multiply and divide by $\frac{\sqrt{5}}{2}$
You will get: $$\sqrt{5}\big(\frac{1}{\sqrt{5}}\sin x + \frac{2}{\sqrt{5}}\cos x\big)$$
which can be expressed in the form of $a(\cos\theta\sin x+\sin\theta\cos x)=a\sin(x+\theta)$
You can express it as $$\sqrt{5}\sin \big(x+\arcsin( \frac{2}{\sqrt{5}})\big)<2$$
which is very easy to solve.
A: Considering equality first.
$$ \sin ( x+ \cos^{-1} \dfrac{1}{\sqrt5})=\dfrac{2}{\sqrt5} ,\; or, \sin ( x+ \tan^{-1} 2)=\frac{2}{\sqrt5}.$$
Solving we get:
x = $2 \pi n+\pi-sin^{-1}(2/sqrt(5))-tan^{-1}(2)$
and x=$2n\pi$
Plotting roughly we get this graph. 
We solve in the domain $0$ to $2\pi$ first as that the period of sine function.
Taking 0 to $\pi-sin^{-1}(2/sqrt(5))-tan^{-1}(2)$ we can understand that $sin(x)+2cos(x)>0$.And in the remaining $\pi-sin^{-1}(2/sqrt(5))-tan^{-1}(2)$ to $2\pi$ $sin(x)+2cos(x)<0$ holds.And that's the answer.
You may generalize the answer as

$2 \pi n-\pi-sin^{-1}(2/sqrt(5))-tan^{-1}(2)<x<2 \pi
 n+sin^{-1}(2/sqrt(5))-tan^{-1}(2)$

A: 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 $\theta = \arcsin \dfrac{1}{\sqrt 5}$
A: The answer is the following set $X$.
$$
X=\{t|2n\pi+2\tan^{-1}(\frac{1}{2})<t<2n\pi+2\pi  {\rm ~for~some~} n \in \mathbb{Z}\}
$$ 

