Mister Exam
-12cos6xsin6x<3 inequation
The solution
You have entered
[src]
-12*cos(6*x)*sin(6*x) < 3
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) < 3$$
sin(6*x)*(-12*cos(6*x)) < 3
Detail solution
Given the inequality:
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) < 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) = 3$$
Solve:
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
This roots
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}$$
=
$$- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) < 3$$
$$\sin{\left(6 \left(- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}\right) \right)} \left(- 12 \cos{\left(6 \left(- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}\right) \right)}\right) < 3$$
one of the solutions of our inequality is:
$$x < - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x > \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3} \wedge x < \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x > \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) < 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) = 3$$
Solve:
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
This roots
$$x_{1} = - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x_{4} = \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3}$$
$$x_{2} = \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x_{3} = \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}$$
=
$$- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(6 x \right)} \left(- 12 \cos{\left(6 x \right)}\right) < 3$$
$$\sin{\left(6 \left(- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}\right) \right)} \left(- 12 \cos{\left(6 \left(- \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3} - \frac{1}{10}\right) \right)}\right) < 3$$
/ / ___________\\ / / ___________\\
|3 | ___ / ___ || |3 | ___ / ___ ||
12*cos|- + 2*atan\-2 + \/ 3 + 2*\/ 2 - \/ 3 /|*sin|- + 2*atan\-2 + \/ 3 + 2*\/ 2 - \/ 3 /| < 3
\5 / \5 /
one of the solutions of our inequality is:
$$x < - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
_____ _____ _____
\ / \ /
-------ο-------ο-------ο-------ο-------
x1 x4 x2 x3Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\operatorname{atan}{\left(-2 + 2 \sqrt{2 - \sqrt{3}} + \sqrt{3} \right)}}{3}$$
$$x > \frac{\operatorname{atan}{\left(- 2 \sqrt{\sqrt{3} + 2} + \sqrt{3} + 2 \right)}}{3} \wedge x < \frac{\operatorname{atan}{\left(- \sqrt{3} + 2 \sqrt{2 - \sqrt{3}} + 2 \right)}}{3}$$
$$x > \frac{\operatorname{atan}{\left(\sqrt{3} + 2 + 2 \sqrt{\sqrt{3} + 2} \right)}}{3}$$
Solving inequality on a graph
Rapid solution
[src]
/ / 7*pi\ / pi 11*pi \\ Or|And|0 <= x, x < ----|, And|x <= --, ----- < x|| \ \ 72 / \ 6 72 //
$$\left(0 \leq x \wedge x < \frac{7 \pi}{72}\right) \vee \left(x \leq \frac{\pi}{6} \wedge \frac{11 \pi}{72} < x\right)$$
((0 <= x)∧(x < 7*pi/72))∨((x <= pi/6)∧(11*pi/72 < x))
Rapid solution 2
[src]
7*pi 11*pi pi
[0, ----) U (-----, --]
72 72 6
$$x\ in\ \left[0, \frac{7 \pi}{72}\right) \cup \left(\frac{11 \pi}{72}, \frac{\pi}{6}\right]$$
x in Union(Interval.Ropen(0, 7*pi/72), Interval.Lopen(11*pi/72, pi/6))