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- Square Inequality Step-by-Step
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How to use it?
- Inequation:
- 0,25*(1/4)*x>=1/64
- а+3<в+2
- 4x+2:2-5x<1
- (1/9)*(x^2)*ln(9-x)-2*ln(9-x)<=0
- Sum of series:
-
1/2
- Derivative of:
-
1/2
- Integral of d{x}:
-
1/2
-
Identical expressions
- sin(x/ four - three)< one / two
- sinus of (x divide by 4 minus 3) less than 1 divide by 2
- sinus of (x divide by four minus three) less than one divide by two
- sinx/4-3<1/2
- sin(x divide by 4-3)<1 divide by 2
-
Similar expressions
- sin(x/4+3)<1/2
- (4x+1)/(2-x)<1/(x+2)
sin(x/4-3)<1/2 inequation
The solution
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[src]
/x \ sin|- - 3| < 1/2 \4 /
$$\sin{\left(\frac{x}{4} - 3 \right)} < \frac{1}{2}$$
sin(x/4 - 3) < 1/2
Detail solution
Given the inequality:
$$\sin{\left(\frac{x}{4} - 3 \right)} < \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{4} - 3 \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{4} - 3 \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{4} - 3 = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}$$
$$\frac{x}{4} - 3 = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi$$
Or
$$\frac{x}{4} - 3 = 2 \pi n + \frac{\pi}{6}$$
$$\frac{x}{4} - 3 = 2 \pi n + \frac{5 \pi}{6}$$
, where n - is a integer
Move
$$-3$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{4} = 2 \pi n + \frac{\pi}{6} + 3$$
$$\frac{x}{4} = 2 \pi n + \frac{5 \pi}{6} + 3$$
Divide both parts of the equation by
$$\frac{1}{4}$$
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
This roots
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(8 \pi n + \frac{2 \pi}{3} + 12\right) + - \frac{1}{10}$$
=
$$8 \pi n + \frac{2 \pi}{3} + \frac{119}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{4} - 3 \right)} < \frac{1}{2}$$
$$\sin{\left(\frac{8 \pi n + \frac{2 \pi}{3} + \frac{119}{10}}{4} - 3 \right)} < \frac{1}{2}$$
one of the solutions of our inequality is:
$$x < 8 \pi n + \frac{2 \pi}{3} + 12$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x > 8 \pi n + \frac{10 \pi}{3} + 12$$
$$\sin{\left(\frac{x}{4} - 3 \right)} < \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{4} - 3 \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{4} - 3 \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{4} - 3 = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}$$
$$\frac{x}{4} - 3 = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi$$
Or
$$\frac{x}{4} - 3 = 2 \pi n + \frac{\pi}{6}$$
$$\frac{x}{4} - 3 = 2 \pi n + \frac{5 \pi}{6}$$
, where n - is a integer
Move
$$-3$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{4} = 2 \pi n + \frac{\pi}{6} + 3$$
$$\frac{x}{4} = 2 \pi n + \frac{5 \pi}{6} + 3$$
Divide both parts of the equation by
$$\frac{1}{4}$$
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
This roots
$$x_{1} = 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x_{2} = 8 \pi n + \frac{10 \pi}{3} + 12$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(8 \pi n + \frac{2 \pi}{3} + 12\right) + - \frac{1}{10}$$
=
$$8 \pi n + \frac{2 \pi}{3} + \frac{119}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{4} - 3 \right)} < \frac{1}{2}$$
$$\sin{\left(\frac{8 \pi n + \frac{2 \pi}{3} + \frac{119}{10}}{4} - 3 \right)} < \frac{1}{2}$$
/ 1 pi \ sin|- -- + -- + 2*pi*n| < 1/2 \ 40 6 /
one of the solutions of our inequality is:
$$x < 8 \pi n + \frac{2 \pi}{3} + 12$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 8 \pi n + \frac{2 \pi}{3} + 12$$
$$x > 8 \pi n + \frac{10 \pi}{3} + 12$$
Solving inequality on a graph
Rapid solution
[src]
/ / / ___ / 2 \ \\ / / ___ / 2 \ \ \\ | | | \/ 3 *\1 + tan (3/2)/ 2*(1 - tan(3/2))*(1 + tan(3/2))|| | | \/ 3 *\1 + tan (3/2)/ 2*(1 - tan(3/2))*(1 + tan(3/2))| || Or|And|0 <= x, x < 8*pi + 8*atan|- -------------------------- + -------------------------------||, And|x <= 8*pi, 8*pi + 8*atan|-------------------------- + -------------------------------| < x|| | | | 2 2 || | | 2 2 | || \ \ \ 1 + tan (3/2) - 4*tan(3/2) 1 + tan (3/2) - 4*tan(3/2) // \ \1 + tan (3/2) - 4*tan(3/2) 1 + tan (3/2) - 4*tan(3/2) / //
$$\left(0 \leq x \wedge x < 8 \operatorname{atan}{\left(\frac{2 \left(1 - \tan{\left(\frac{3}{2} \right)}\right) \left(1 + \tan{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} - \frac{\sqrt{3} \left(1 + \tan^{2}{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} \right)} + 8 \pi\right) \vee \left(x \leq 8 \pi \wedge 8 \operatorname{atan}{\left(\frac{2 \left(1 - \tan{\left(\frac{3}{2} \right)}\right) \left(1 + \tan{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} + \frac{\sqrt{3} \left(1 + \tan^{2}{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} \right)} + 8 \pi < x\right)$$
((0 <= x)∧(x < 8*pi + 8*atan(-sqrt(3)*(1 + tan(3/2)^2)/(1 + tan(3/2)^2 - 4*tan(3/2)) + 2*(1 - tan(3/2))*(1 + tan(3/2))/(1 + tan(3/2)^2 - 4*tan(3/2)))))∨((x <= 8*pi)∧(8*pi + 8*atan(sqrt(3)*(1 + tan(3/2)^2)/(1 + tan(3/2)^2 - 4*tan(3/2)) + 2*(1 - tan(3/2))*(1 + tan(3/2))/(1 + tan(3/2)^2 - 4*tan(3/2))) < x))
Rapid solution 2
[src]
/ ___ / 2 \ \ / ___ / 2 \ \
| \/ 3 *\1 + tan (3/2)/ 2*(1 - tan(3/2))*(1 + tan(3/2))| | \/ 3 *\1 + tan (3/2)/ 2*(1 - tan(3/2))*(1 + tan(3/2))|
[0, 8*pi + 8*atan|- -------------------------- + -------------------------------|) U (8*pi + 8*atan|-------------------------- + -------------------------------|, 8*pi]
| 2 2 | | 2 2 |
\ 1 + tan (3/2) - 4*tan(3/2) 1 + tan (3/2) - 4*tan(3/2) / \1 + tan (3/2) - 4*tan(3/2) 1 + tan (3/2) - 4*tan(3/2) /
$$x\ in\ \left[0, 8 \operatorname{atan}{\left(\frac{2 \left(1 - \tan{\left(\frac{3}{2} \right)}\right) \left(1 + \tan{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} - \frac{\sqrt{3} \left(1 + \tan^{2}{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} \right)} + 8 \pi\right) \cup \left(8 \operatorname{atan}{\left(\frac{2 \left(1 - \tan{\left(\frac{3}{2} \right)}\right) \left(1 + \tan{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} + \frac{\sqrt{3} \left(1 + \tan^{2}{\left(\frac{3}{2} \right)}\right)}{- 4 \tan{\left(\frac{3}{2} \right)} + 1 + \tan^{2}{\left(\frac{3}{2} \right)}} \right)} + 8 \pi, 8 \pi\right]$$
x in Union(Interval.Ropen(0, 8*atan(2*(1 - tan(3/2))*(1 + tan(3/2))/(-4*tan(3/2) + 1 + tan(3/2)^2) - sqrt(3)*(1 + tan(3/2)^2)/(-4*tan(3/2) + 1 + tan(3/2)^2)) + 8*pi), Interval.Lopen(8*atan(2*(1 - tan(3/2))*(1 + tan(3/2))/(-4*tan(3/2) + 1 + tan(3/2)^2) + sqrt(3)*(1 + tan(3/2)^2)/(-4*tan(3/2) + 1 + tan(3/2)^2)) + 8*pi, 8*pi))