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Inequation:
(x-4)^2>16-x^2
16^x-9<1
5^(1-x)>0,2
(|x+3|)+4*x>=6
Identical expressions
sin4x/ three >-sqrt(three)/ two
sinus of 4x divide by 3 greater than minus square root of (3) divide by 2
sinus of 4x divide by three greater than minus square root of (three) divide by two
sin4x/3>-√(3)/2
sin4x/3>-sqrt3/2
sin4x divide by 3>-sqrt(3) divide by 2
Similar expressions
sin(4*x)/3>-3/2
sin(4*x)/3>=sqrt(3)/2
sin((4*x)/3)>-sqrt(3)*1/2
sin(x/4)<-sqrt3/2
sin(4*x/3)>-(sqrt(3)/2)
cos>=-sqrt3/2
sin4x/3>+sqrt(3)/2
cos(2x)<=-sqrt(3/2)
cosx>=-sqrt3/2

Solving inequalities/ sin4x/3>-sqrt(3)/2

sin4x/3>-sqrt(3)/2 inequation

The solution

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              ___ 
sin(4*x)   -\/ 3  
-------- > -------
   3          2

\frac{\sin{\left(4 x \right)}}{3} > \frac{\left(-1\right) \sqrt{3}}{2}

sin(4*x)/3 > -sqrt(3)/2

Detail solution

Given the inequality:

\frac{\sin{\left(4 x \right)}}{3} > \frac{\left(-1\right) \sqrt{3}}{2}

To solve this inequality, we must first solve the corresponding equation:

\frac{\sin{\left(4 x \right)}}{3} = \frac{\left(-1\right) \sqrt{3}}{2}

Solve:
Given the equation

\frac{\sin{\left(4 x \right)}}{3} = \frac{\left(-1\right) \sqrt{3}}{2}

- this is the simplest trigonometric equation

Divide both parts of the equation by 1/3

The equation is transformed to

\sin{\left(4 x \right)} = - \frac{3 \sqrt{3}}{2}

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{\pi}{4} + \frac{\operatorname{asin}{\left(\frac{3 \sqrt{3}}{2} \right)}}{4}

x_{2} = - \frac{\operatorname{asin}{\left(\frac{3 \sqrt{3}}{2} \right)}}{4}

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

\frac{\sin{\left(4 \cdot 0 \right)}}{3} > \frac{\left(-1\right) \sqrt{3}}{2}

       ___ 
    -\/ 3  
0 > -------
       2

so the inequality is always executed

Solving inequality on a graph

Rapid solution 2 [src]

(-oo, oo)

x\ in\ \left(-\infty, \infty\right)

x in Interval(-oo, oo)

Rapid solution [src]

And(-oo < x, x < oo)

-\infty < x \wedge x < \infty

(-oo < x)∧(x < oo)

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Identical expressions

Similar expressions

sin4x/3>-sqrt(3)/2 inequation

The solution