sin(x)+sqrt(3)*cos(x)>sqrt(2) inequation

Given the inequality:
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} > \sqrt{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} = \sqrt{2}$$
Solve:
$$x_{1} = 2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
$$x_{2} = 2 \operatorname{atan}{\left(\frac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
$$x_{1} = 2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
$$x_{2} = 2 \operatorname{atan}{\left(\frac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
This roots
$$x_{1} = 2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
$$x_{2} = 2 \operatorname{atan}{\left(\frac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)} - \frac{1}{10}$$
=
$$2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x \right)} + \sqrt{3} \cos{\left(x \right)} > \sqrt{2}$$
$$\sin{\left(2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)} - \frac{1}{10} \right)} + \sqrt{3} \cos{\left(2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)} - \frac{1}{10} \right)} > \sqrt{2}$$

         /             /        ___  \\      /             /        ___  \\        
  ___    |  1          |  1 - \/ 2   ||      |  1          |  1 - \/ 2   ||     ___
\/ 3 *cos|- -- + 2*atan|-------------|| + sin|- -- + 2*atan|-------------|| > \/ 2 
         |  10         |  ___     ___||      |  10         |  ___     ___||   
         \             \\/ 2  + \/ 3 //      \             \\/ 2  + \/ 3 //        

Then
$$x < 2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$
no execute
one of the solutions of our inequality is:
$$x > 2 \operatorname{atan}{\left(\frac{1 - \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)} \wedge x < 2 \operatorname{atan}{\left(\frac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}} \right)}$$

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       x1      x2