Mister Exam
tan(2*x+pi/6)>=sqrt(3) inequation
The solution
You have entered
[src]
/ pi\ ___ tan|2*x + --| >= \/ 3 \ 6 /
$$\tan{\left(2 x + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
tan(2*x + pi/6) >= sqrt(3)
Detail solution
Given the inequality:
$$\tan{\left(2 x + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{6} = \pi n + \operatorname{atan}{\left(\sqrt{3} \right)}$$
Or
$$2 x + \frac{\pi}{6} = \pi n + \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = \pi n + \frac{\pi}{6}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\tan{\left(2 x + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
$$\tan{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}\right) + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
but
Then
$$x \leq \frac{\pi n}{2} + \frac{\pi}{12}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi n}{2} + \frac{\pi}{12}$$
$$\tan{\left(2 x + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{6} = \pi n + \operatorname{atan}{\left(\sqrt{3} \right)}$$
Or
$$2 x + \frac{\pi}{6} = \pi n + \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = \pi n + \frac{\pi}{6}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\tan{\left(2 x + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
$$\tan{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}\right) + \frac{\pi}{6} \right)} \geq \sqrt{3}$$
/ 1 pi \ ___ tan|- - + -- + pi*n| >= \/ 3 \ 5 3 /
but
/ 1 pi \ ___ tan|- - + -- + pi*n| < \/ 3 \ 5 3 /
Then
$$x \leq \frac{\pi n}{2} + \frac{\pi}{12}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi n}{2} + \frac{\pi}{12}$$
_____
/
-------•-------
x1
Solving inequality on a graph
Rapid solution
[src]
/ / ___ ___\ \ | pi |\/ 2 - \/ 6 | | And|x <= --, -atan|-------------| <= x| | 6 | ___ ___| | \ \\/ 2 + \/ 6 / /
$$x \leq \frac{\pi}{6} \wedge - \operatorname{atan}{\left(\frac{- \sqrt{6} + \sqrt{2}}{\sqrt{2} + \sqrt{6}} \right)} \leq x$$
(x <= pi/6)∧(-atan((sqrt(2) - sqrt(6))/(sqrt(2) + sqrt(6))) <= x)
Rapid solution 2
[src]
/ ___ ___\
|\/ 2 - \/ 6 | pi
[-atan|-------------|, --]
| ___ ___| 6
\\/ 2 + \/ 6 /
$$x\ in\ \left[- \operatorname{atan}{\left(\frac{- \sqrt{6} + \sqrt{2}}{\sqrt{2} + \sqrt{6}} \right)}, \frac{\pi}{6}\right]$$
x in Interval(-atan((-sqrt(6) + sqrt(2))/(sqrt(2) + sqrt(6))), pi/6)