Given the inequality:
$$\sin{\left(3 x + 1 \right)} < \frac{2^{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(3 x + 1 \right)} = \frac{2^{2}}{2}$$
Solve:
Given the equation
$$\sin{\left(3 x + 1 \right)} = \frac{2^{2}}{2}$$
- this is the simplest trigonometric equation
As right part of the equation
modulo =
True
but sin
can no be more than 1 or less than -1
so the solution of the equation d'not exist.
$$x_{1} = - \frac{1}{3} + \frac{\operatorname{asin}{\left(2 \right)}}{3}$$
$$x_{2} = - \frac{1}{3} + \frac{\pi}{3} - \frac{\operatorname{asin}{\left(2 \right)}}{3}$$
Exclude the complex solutions:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example
x0 = 0
$$\sin{\left(0 \cdot 3 + 1 \right)} < \frac{2^{2}}{2}$$
sin(1) < 2
so the inequality is always executed