sin2x+sqrt3*cos2x>=-1 inequation

Given the inequality:
$$\sin{\left(2 x \right)} + \sqrt{3} \cos{\left(2 x \right)} \geq -1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x \right)} + \sqrt{3} \cos{\left(2 x \right)} = -1$$
Solve:
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{5 \pi}{12}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{5 \pi}{12}$$
This roots
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{5 \pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(2 x \right)} + \sqrt{3} \cos{\left(2 x \right)} \geq -1$$
$$\sin{\left(2 \left(- \frac{\pi}{4} - \frac{1}{10}\right) \right)} + \sqrt{3} \cos{\left(2 \left(- \frac{\pi}{4} - \frac{1}{10}\right) \right)} \geq -1$$

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-cos(1/5) - \/ 3 *sin(1/5) >= -1
      

but

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-cos(1/5) - \/ 3 *sin(1/5) < -1
     

Then
$$x \leq - \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \frac{\pi}{4} \wedge x \leq \frac{5 \pi}{12}$$

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       x1      x2