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sin(3x+1)<(sqr(2)/2) inequation

A inequation with variable

The solution

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                2
               2 
sin(3*x + 1) < --
               2 
$$\sin{\left(3 x + 1 \right)} < \frac{2^{2}}{2}$$
sin(3*x + 1) < 2^2/2
Detail solution
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
Solving inequality on a graph
Rapid solution [src]
And(-oo < x, x < oo)
$$-\infty < x \wedge x < \infty$$
(-oo < x)∧(x < oo)
Rapid solution 2 [src]
(-oo, oo)
$$x\ in\ \left(-\infty, \infty\right)$$
x in Interval(-oo, oo)