Given the inequality:
$$\sin{\left(x - \frac{\pi}{3} \right)} \geq \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x - \frac{\pi}{3} \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(x - \frac{\pi}{3} \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
Divide both parts of the equation by $-1$
The equation is transformed to
$$\cos{\left(x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
This equation is transformed to
$$x + \frac{\pi}{6} = 2 \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$x + \frac{\pi}{6} = 2 \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$x + \frac{\pi}{6} = 2 \pi n + \frac{2 \pi}{3}$$
$$x + \frac{\pi}{6} = 2 \pi n - \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation with the opposite sign, in total:
$$x = 2 \pi n + \frac{\pi}{2}$$
$$x = 2 \pi n - \frac{\pi}{2}$$
$$x_{1} = 2 \pi n + \frac{\pi}{2}$$
$$x_{2} = 2 \pi n - \frac{\pi}{2}$$
$$x_{1} = 2 \pi n + \frac{\pi}{2}$$
$$x_{2} = 2 \pi n - \frac{\pi}{2}$$
This roots
$$x_{1} = 2 \pi n + \frac{\pi}{2}$$
$$x_{2} = 2 \pi n - \frac{\pi}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n + \frac{\pi}{2}\right) - \frac{1}{10}$$
=
$$2 \pi n - \frac{1}{10} + \frac{\pi}{2}$$
substitute to the expression
$$\sin{\left(x - \frac{\pi}{3} \right)} \geq \frac{1}{2}$$
$$\sin{\left(\left(2 \pi n - \frac{1}{10} + \frac{\pi}{2}\right) - \frac{\pi}{3} \right)} \geq \frac{1}{2}$$
/1 pi\
cos|-- + --| >= 1/2
\10 3 /
but
/1 pi\
cos|-- + --| < 1/2
\10 3 /
Then
$$x \leq 2 \pi n + \frac{\pi}{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq 2 \pi n + \frac{\pi}{2} \wedge x \leq 2 \pi n - \frac{\pi}{2}$$
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x_1 x_2