Mister Exam
sin(x/6)>-sqrt(3)/2 inequation
The solution
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___ /x\ -\/ 3 sin|-| > ------- \6/ 2
$$\sin{\left(\frac{x}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
sin(x/6) > (-sqrt(3))/2
Detail solution
Given the inequality:
$$\sin{\left(\frac{x}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{6} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{6} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{6} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$\frac{x}{6} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$\frac{x}{6} = 2 \pi n - \frac{\pi}{3}$$
$$\frac{x}{6} = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{6}$$
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
This roots
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(12 \pi n - 2 \pi\right) + - \frac{1}{10}$$
=
$$12 \pi n - 2 \pi - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\sin{\left(\frac{12 \pi n - 2 \pi - \frac{1}{10}}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
Then
$$x < 12 \pi n - 2 \pi$$
no execute
one of the solutions of our inequality is:
$$x > 12 \pi n - 2 \pi \wedge x < 12 \pi n + 8 \pi$$
$$\sin{\left(\frac{x}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{6} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{6} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{6} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$\frac{x}{6} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$\frac{x}{6} = 2 \pi n - \frac{\pi}{3}$$
$$\frac{x}{6} = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{6}$$
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
This roots
$$x_{1} = 12 \pi n - 2 \pi$$
$$x_{2} = 12 \pi n + 8 \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(12 \pi n - 2 \pi\right) + - \frac{1}{10}$$
=
$$12 \pi n - 2 \pi - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\sin{\left(\frac{12 \pi n - 2 \pi - \frac{1}{10}}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
___
/1 pi \ -\/ 3
-sin|-- + -- - 2*pi*n| > -------
\60 3 / 2
Then
$$x < 12 \pi n - 2 \pi$$
no execute
one of the solutions of our inequality is:
$$x > 12 \pi n - 2 \pi \wedge x < 12 \pi n + 8 \pi$$
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/ \
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x1 x2
Solving inequality on a graph
Rapid solution 2
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[0, 8*pi) U (10*pi, 12*pi]
$$x\ in\ \left[0, 8 \pi\right) \cup \left(10 \pi, 12 \pi\right]$$
x in Union(Interval.Ropen(0, 8*pi), Interval.Lopen(10*pi, 12*pi))
Rapid solution
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Or(And(0 <= x, x < 8*pi), And(x <= 12*pi, 10*pi < x))
$$\left(0 \leq x \wedge x < 8 \pi\right) \vee \left(x \leq 12 \pi \wedge 10 \pi < x\right)$$
((0 <= x)∧(x < 8*pi))∨((x <= 12*pi)∧(10*pi < x))