sinx-3cosx<0 inequation

Given the inequality:
$$\sin{\left(x \right)} - 3 \cos{\left(x \right)} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x \right)} - 3 \cos{\left(x \right)} = 0$$
Solve:
Given the equation
$$\sin{\left(x \right)} - 3 \cos{\left(x \right)} = 0$$
transform:
$$\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} = 3$$
or
$$\tan{\left(x \right)} = 3$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(-3 \right)}$$
Or
$$x = \pi n - \operatorname{atan}{\left(3 \right)}$$
, where n - is a integer
$$x_{1} = \pi n - \operatorname{atan}{\left(3 \right)}$$
$$x_{1} = \pi n - \operatorname{atan}{\left(3 \right)}$$
This roots
$$x_{1} = \pi n - \operatorname{atan}{\left(3 \right)}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \operatorname{atan}{\left(3 \right)}\right) + - \frac{1}{10}$$
=
$$\pi n - \operatorname{atan}{\left(3 \right)} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x \right)} - 3 \cos{\left(x \right)} < 0$$
$$\sin{\left(\pi n - \operatorname{atan}{\left(3 \right)} - \frac{1}{10} \right)} - 3 \cos{\left(\pi n - \operatorname{atan}{\left(3 \right)} - \frac{1}{10} \right)} < 0$$

-sin(1/10 - pi*n + atan(3)) - 3*cos(1/10 - pi*n + atan(3)) < 0

but

-sin(1/10 - pi*n + atan(3)) - 3*cos(1/10 - pi*n + atan(3)) > 0

Then
$$x < \pi n - \operatorname{atan}{\left(3 \right)}$$
no execute
the solution of our inequality is:
$$x > \pi n - \operatorname{atan}{\left(3 \right)}$$

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