Mister Exam
sin(x/4-3)>0 inequation
The solution
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/x \ sin|- - 3| > 0 \4 /
$$\sin{\left(\frac{x}{4} - 3 \right)} > 0$$
sin(x/4 - 3) > 0
Detail solution
Given the inequality:
$$\sin{\left(\frac{x}{4} - 3 \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
- this is the simplest trigonometric equation
We get:
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
This equation is transformed to
$$\frac{x}{4} - 3 = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$\frac{x}{4} - 3 = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$\frac{x}{4} - 3 = 2 \pi n$$
$$\frac{x}{4} - 3 = 2 \pi n + \pi$$
, where n - is a integer
Move
$$-3$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{4} = 2 \pi n + 3$$
$$\frac{x}{4} = 2 \pi n + 3 + \pi$$
Divide both parts of the equation by
$$\frac{1}{4}$$
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
This roots
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(8 \pi n + 12\right) + - \frac{1}{10}$$
=
$$8 \pi n + \frac{119}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{4} - 3 \right)} > 0$$
$$\sin{\left(\frac{8 \pi n + \frac{119}{10}}{4} - 3 \right)} > 0$$
Then
$$x < 8 \pi n + 12$$
no execute
one of the solutions of our inequality is:
$$x > 8 \pi n + 12 \wedge x < 8 \pi n + 12 + 4 \pi$$
$$\sin{\left(\frac{x}{4} - 3 \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
- this is the simplest trigonometric equation
with the change of sign in 0
We get:
$$\sin{\left(\frac{x}{4} - 3 \right)} = 0$$
This equation is transformed to
$$\frac{x}{4} - 3 = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$\frac{x}{4} - 3 = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$\frac{x}{4} - 3 = 2 \pi n$$
$$\frac{x}{4} - 3 = 2 \pi n + \pi$$
, where n - is a integer
Move
$$-3$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{4} = 2 \pi n + 3$$
$$\frac{x}{4} = 2 \pi n + 3 + \pi$$
Divide both parts of the equation by
$$\frac{1}{4}$$
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
This roots
$$x_{1} = 8 \pi n + 12$$
$$x_{2} = 8 \pi n + 12 + 4 \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(8 \pi n + 12\right) + - \frac{1}{10}$$
=
$$8 \pi n + \frac{119}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{4} - 3 \right)} > 0$$
$$\sin{\left(\frac{8 \pi n + \frac{119}{10}}{4} - 3 \right)} > 0$$
sin(-1/40 + 2*pi*n) > 0
Then
$$x < 8 \pi n + 12$$
no execute
one of the solutions of our inequality is:
$$x > 8 \pi n + 12 \wedge x < 8 \pi n + 12 + 4 \pi$$
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x1 x2
Solving inequality on a graph
Rapid solution 2
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/ 1 \
(12, - 8*atan|--------| + 8*pi)
\tan(3/2)/
$$x\ in\ \left(12, - 8 \operatorname{atan}{\left(\frac{1}{\tan{\left(\frac{3}{2} \right)}} \right)} + 8 \pi\right)$$
x in Interval.open(12, -8*atan(1/tan(3/2)) + 8*pi)
Rapid solution
[src]
/ / 1 \ \ And|12 < x, x < - 8*atan|--------| + 8*pi| \ \tan(3/2)/ /
$$12 < x \wedge x < - 8 \operatorname{atan}{\left(\frac{1}{\tan{\left(\frac{3}{2} \right)}} \right)} + 8 \pi$$
(12 < x)∧(x < -8*atan(1/tan(3/2)) + 8*pi)