sin(x)+sqrt(3)cos(x)<1 inequation

Given the inequality:
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} = 1$$
Solve:
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{\pi}{2}$$
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{\pi}{2}$$
This roots
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{\pi}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{6} - \frac{1}{10}$$
=
$$- \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} < 1$$
$$\sin{\left(- \frac{\pi}{6} - \frac{1}{10} \right)} + \sqrt{3} \cos{\left(- \frac{\pi}{6} - \frac{1}{10} \right)} < 1$$

     /1    pi\     ___    /1    pi\    
- sin|-- + --| + \/ 3 *cos|-- + --| < 1
     \10   6 /            \10   6 /    

one of the solutions of our inequality is:
$$x < - \frac{\pi}{6}$$

 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\pi}{6}$$
$$x > \frac{\pi}{2}$$