Given the inequality:
$$\sin{\left(3 x + 1 \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(3 x + 1 \right)} = 0$$
Solve:
Given the equation
$$\sin{\left(3 x + 1 \right)} = 0$$
- this is the simplest trigonometric equation
with the change of sign in 0
We get:
$$\sin{\left(3 x + 1 \right)} = 0$$
This equation is transformed to
$$3 x + 1 = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$3 x + 1 = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$3 x + 1 = 2 \pi n$$
$$3 x + 1 = 2 \pi n + \pi$$
, where n - is a integer
Move
$$1$$
to right part of the equation
with the opposite sign, in total:
$$3 x = 2 \pi n - 1$$
$$3 x = 2 \pi n - 1 + \pi$$
Divide both parts of the equation by
$$3$$
$$x_{1} = \frac{2 \pi n}{3} - \frac{1}{3}$$
$$x_{2} = \frac{2 \pi n}{3} - \frac{1}{3} + \frac{\pi}{3}$$
$$x_{1} = \frac{2 \pi n}{3} - \frac{1}{3}$$
$$x_{2} = \frac{2 \pi n}{3} - \frac{1}{3} + \frac{\pi}{3}$$
This roots
$$x_{1} = \frac{2 \pi n}{3} - \frac{1}{3}$$
$$x_{2} = \frac{2 \pi n}{3} - \frac{1}{3} + \frac{\pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{2 \pi n}{3} - \frac{1}{3}\right) + - \frac{1}{10}$$
=
$$\frac{2 \pi n}{3} - \frac{13}{30}$$
substitute to the expression
$$\sin{\left(3 x + 1 \right)} > 0$$
$$\sin{\left(3 \left(\frac{2 \pi n}{3} - \frac{13}{30}\right) + 1 \right)} > 0$$
sin(-3/10 + 2*pi*n) > 0
Then
$$x < \frac{2 \pi n}{3} - \frac{1}{3}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{2 \pi n}{3} - \frac{1}{3} \wedge x < \frac{2 \pi n}{3} - \frac{1}{3} + \frac{\pi}{3}$$
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