Mister Exam
2*sin6x+sqrt(3)<0 inequation
The solution
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___ 2*sin(6*x) + \/ 3 < 0
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
2*sin(6*x) + sqrt(3) < 0
Detail solution
Given the inequality:
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
Solve:
Given the equation
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
- this is the simplest trigonometric equation
We get:
$$2 \sin{\left(6 x \right)} = - \sqrt{3}$$
The equation is transformed to
$$\sin{\left(6 x \right)} = - \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$6 x = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$6 x = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$6 x = 2 \pi n - \frac{\pi}{3}$$
$$6 x = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
This roots
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} - \frac{\pi}{18}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}$$
substitute to the expression
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
$$2 \sin{\left(6 \left(\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}\right) \right)} + \sqrt{3} < 0$$
one of the solutions of our inequality is:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x > \frac{\pi n}{3} + \frac{2 \pi}{9}$$
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
Solve:
Given the equation
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
- this is the simplest trigonometric equation
Move sqrt(3) to right part of the equation
with the change of sign in sqrt(3)
We get:
$$2 \sin{\left(6 x \right)} = - \sqrt{3}$$
Divide both parts of the equation by 2
The equation is transformed to
$$\sin{\left(6 x \right)} = - \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$6 x = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$6 x = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$6 x = 2 \pi n - \frac{\pi}{3}$$
$$6 x = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
This roots
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} - \frac{\pi}{18}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}$$
substitute to the expression
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
$$2 \sin{\left(6 \left(\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}\right) \right)} + \sqrt{3} < 0$$
___ /3 pi \
\/ 3 - 2*sin|- + -- - 2*pi*n| < 0
\5 3 / one of the solutions of our inequality is:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x > \frac{\pi n}{3} + \frac{2 \pi}{9}$$
Solving inequality on a graph
Rapid solution
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/2*pi 5*pi\ And|---- < x, x < ----| \ 9 18 /
$$\frac{2 \pi}{9} < x \wedge x < \frac{5 \pi}{18}$$
(2*pi/9 < x)∧(x < 5*pi/18)
Rapid solution 2
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2*pi 5*pi (----, ----) 9 18
$$x\ in\ \left(\frac{2 \pi}{9}, \frac{5 \pi}{18}\right)$$
x in Interval.open(2*pi/9, 5*pi/18)