Mister Exam
-2*sin2x>-sqrt(3) inequation
The solution
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___ -2*sin(2*x) > -\/ 3
$$- 2 \sin{\left(2 x \right)} > - \sqrt{3}$$
-2*sin(2*x) > -sqrt(3)
Detail solution
Given the inequality:
$$- 2 \sin{\left(2 x \right)} > - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2 \sin{\left(2 x \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$- 2 \sin{\left(2 x \right)} = - \sqrt{3}$$
- this is the simplest trigonometric equation
The equation is transformed to
$$\sin{\left(2 x \right)} = \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$2 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$2 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$2 x = 2 \pi n + \frac{\pi}{3}$$
$$2 x = 2 \pi n + \frac{2 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$- 2 \sin{\left(2 x \right)} > - \sqrt{3}$$
$$- 2 \sin{\left(2 \left(\pi n - \frac{1}{10} + \frac{\pi}{6}\right) \right)} > - \sqrt{3}$$
one of the solutions of our inequality is:
$$x < \pi n + \frac{\pi}{6}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n + \frac{\pi}{6}$$
$$x > \pi n + \frac{\pi}{3}$$
$$- 2 \sin{\left(2 x \right)} > - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2 \sin{\left(2 x \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$- 2 \sin{\left(2 x \right)} = - \sqrt{3}$$
- this is the simplest trigonometric equation
Divide both parts of the equation by -2
The equation is transformed to
$$\sin{\left(2 x \right)} = \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$2 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$2 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$2 x = 2 \pi n + \frac{\pi}{3}$$
$$2 x = 2 \pi n + \frac{2 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{\pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$- 2 \sin{\left(2 x \right)} > - \sqrt{3}$$
$$- 2 \sin{\left(2 \left(\pi n - \frac{1}{10} + \frac{\pi}{6}\right) \right)} > - \sqrt{3}$$
/ 1 pi \ ___
-2*sin|- - + -- + 2*pi*n| > -\/ 3
\ 5 3 / one of the solutions of our inequality is:
$$x < \pi n + \frac{\pi}{6}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n + \frac{\pi}{6}$$
$$x > \pi n + \frac{\pi}{3}$$
Solving inequality on a graph
Rapid solution
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/ / pi\ / pi \\ Or|And|0 <= x, x < --|, And|x <= pi, -- < x|| \ \ 6 / \ 3 //
$$\left(0 \leq x \wedge x < \frac{\pi}{6}\right) \vee \left(x \leq \pi \wedge \frac{\pi}{3} < x\right)$$
((0 <= x)∧(x < pi/6))∨((x <= pi)∧(pi/3 < x))
Rapid solution 2
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pi pi
[0, --) U (--, pi]
6 3
$$x\ in\ \left[0, \frac{\pi}{6}\right) \cup \left(\frac{\pi}{3}, \pi\right]$$
x in Union(Interval.Ropen(0, pi/6), Interval.Lopen(pi/3, pi))