Mister Exam
ctg(2x+pi)≥-sqrt(3) inequation
The solution
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___ cot(2*x + pi) >= -\/ 3
$$\cot{\left(2 x + \pi \right)} \geq - \sqrt{3}$$
cot(2*x + pi) >= -sqrt(3)
Detail solution
Given the inequality:
$$\cot{\left(2 x + \pi \right)} \geq - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(2 x + \pi \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\cot{\left(2 x + \pi \right)} = - \sqrt{3}$$
transform
$$\cot{\left(2 x \right)} + \sqrt{3} = 0$$
$$\cot{\left(2 x \right)} + \sqrt{3} = 0$$
Do replacement
$$w = \cot{\left(2 x \right)}$$
Expand brackets in the left part
Divide both parts of the equation by (w + sqrt(3))/w
We get the answer: w = -sqrt(3)
do backward replacement
$$\cot{\left(2 x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{12}$$
$$x_{1} = - \frac{\pi}{12}$$
This roots
$$x_{1} = - \frac{\pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{12} - \frac{1}{10}$$
=
$$- \frac{\pi}{12} - \frac{1}{10}$$
substitute to the expression
$$\cot{\left(2 x + \pi \right)} \geq - \sqrt{3}$$
$$\cot{\left(2 \left(- \frac{\pi}{12} - \frac{1}{10}\right) + \pi \right)} \geq - \sqrt{3}$$
the solution of our inequality is:
$$x \leq - \frac{\pi}{12}$$
$$\cot{\left(2 x + \pi \right)} \geq - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(2 x + \pi \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\cot{\left(2 x + \pi \right)} = - \sqrt{3}$$
transform
$$\cot{\left(2 x \right)} + \sqrt{3} = 0$$
$$\cot{\left(2 x \right)} + \sqrt{3} = 0$$
Do replacement
$$w = \cot{\left(2 x \right)}$$
Expand brackets in the left part
w + sqrt3 = 0
Divide both parts of the equation by (w + sqrt(3))/w
w = 0 / ((w + sqrt(3))/w)
We get the answer: w = -sqrt(3)
do backward replacement
$$\cot{\left(2 x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{12}$$
$$x_{1} = - \frac{\pi}{12}$$
This roots
$$x_{1} = - \frac{\pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{12} - \frac{1}{10}$$
=
$$- \frac{\pi}{12} - \frac{1}{10}$$
substitute to the expression
$$\cot{\left(2 x + \pi \right)} \geq - \sqrt{3}$$
$$\cot{\left(2 \left(- \frac{\pi}{12} - \frac{1}{10}\right) + \pi \right)} \geq - \sqrt{3}$$
/1 pi\ ___
-cot|- + --| >= -\/ 3
\5 6 / the solution of our inequality is:
$$x \leq - \frac{\pi}{12}$$
_____
\
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x1
Rapid solution 2
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/ ___ ___\
|\/ 2 + \/ 6 |
(0, -atan|-------------|]
| ___ ___|
\\/ 2 - \/ 6 /
$$x\ in\ \left(0, - \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)}\right]$$
x in Interval.Lopen(0, -atan((sqrt(2) + sqrt(6))/(-sqrt(6) + sqrt(2))))
Rapid solution
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/ / ___ ___\ \ | |\/ 2 + \/ 6 | | And|x <= -atan|-------------|, 0 < x| | | ___ ___| | \ \\/ 2 - \/ 6 / /
$$x \leq - \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)} \wedge 0 < x$$
(0 < x)∧(x <= -atan((sqrt(2) + sqrt(6))/(sqrt(2) - sqrt(6))))