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.

$\sin x+2\cos x=2(\frac{1}{2}\sin x + \cos x)$
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.
