Mister Exam
tg(x/2+П/12)<=-sqrt(3) inequation
The solution
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/x pi\ ___ tan|- + --| <= -\/ 3 \2 12/
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
tan(x/2 + pi/12) <= -sqrt(3)
Detail solution
Given the inequality:
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} = - \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{2} + \frac{\pi}{12} = \pi n + \operatorname{atan}{\left(- \sqrt{3} \right)}$$
Or
$$\frac{x}{2} + \frac{\pi}{12} = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{12}$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{2} = \pi n - \frac{5 \pi}{12}$$
Divide both parts of the equation by
$$\frac{1}{2}$$
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
This roots
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{5 \pi}{6}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{5 \pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
$$\tan{\left(\frac{2 \pi n - \frac{5 \pi}{6} - \frac{1}{10}}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
the solution of our inequality is:
$$x \leq 2 \pi n - \frac{5 \pi}{6}$$
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} = - \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{2} + \frac{\pi}{12} = \pi n + \operatorname{atan}{\left(- \sqrt{3} \right)}$$
Or
$$\frac{x}{2} + \frac{\pi}{12} = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{12}$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{2} = \pi n - \frac{5 \pi}{12}$$
Divide both parts of the equation by
$$\frac{1}{2}$$
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
This roots
$$x_{1} = 2 \pi n - \frac{5 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{5 \pi}{6}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{5 \pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(\frac{x}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
$$\tan{\left(\frac{2 \pi n - \frac{5 \pi}{6} - \frac{1}{10}}{2} + \frac{\pi}{12} \right)} \leq - \sqrt{3}$$
/1 pi \ ___
-tan|-- + -- - pi*n| <= -\/ 3
\20 3 / the solution of our inequality is:
$$x \leq 2 \pi n - \frac{5 \pi}{6}$$
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