Mister Exam
2cos((𝑥/3)+2)≥1 inequation
The solution
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/x \
2*cos|- + 2| >= 1
\3 /
$$2 \cos{\left(\frac{x}{3} + 2 \right)} \geq 1$$
2*cos(x/3 + 2) >= 1
Detail solution
Given the inequality:
$$2 \cos{\left(\frac{x}{3} + 2 \right)} \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(\frac{x}{3} + 2 \right)} = 1$$
Solve:
Given the equation
$$2 \cos{\left(\frac{x}{3} + 2 \right)} = 1$$
- this is the simplest trigonometric equation
The equation is transformed to
$$\cos{\left(\frac{x}{3} + 2 \right)} = \frac{1}{2}$$
This equation is transformed to
$$\frac{x}{3} + 2 = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$\frac{x}{3} + 2 = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$\frac{x}{3} + 2 = \pi n + \frac{\pi}{3}$$
$$\frac{x}{3} + 2 = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
Move
$$2$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{3} = \pi n - 2 + \frac{\pi}{3}$$
$$\frac{x}{3} = \pi n - \frac{2 \pi}{3} - 2$$
Divide both parts of the equation by
$$\frac{1}{3}$$
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
This roots
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(3 \pi n - 6 + \pi\right) + - \frac{1}{10}$$
=
$$3 \pi n - \frac{61}{10} + \pi$$
substitute to the expression
$$2 \cos{\left(\frac{x}{3} + 2 \right)} \geq 1$$
$$2 \cos{\left(\frac{3 \pi n - \frac{61}{10} + \pi}{3} + 2 \right)} \geq 1$$
but
Then
$$x \leq 3 \pi n - 6 + \pi$$
no execute
one of the solutions of our inequality is:
$$x \geq 3 \pi n - 6 + \pi \wedge x \leq 3 \pi n - 2 \pi - 6$$
$$2 \cos{\left(\frac{x}{3} + 2 \right)} \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(\frac{x}{3} + 2 \right)} = 1$$
Solve:
Given the equation
$$2 \cos{\left(\frac{x}{3} + 2 \right)} = 1$$
- this is the simplest trigonometric equation
Divide both parts of the equation by 2
The equation is transformed to
$$\cos{\left(\frac{x}{3} + 2 \right)} = \frac{1}{2}$$
This equation is transformed to
$$\frac{x}{3} + 2 = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$\frac{x}{3} + 2 = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$\frac{x}{3} + 2 = \pi n + \frac{\pi}{3}$$
$$\frac{x}{3} + 2 = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
Move
$$2$$
to right part of the equation
with the opposite sign, in total:
$$\frac{x}{3} = \pi n - 2 + \frac{\pi}{3}$$
$$\frac{x}{3} = \pi n - \frac{2 \pi}{3} - 2$$
Divide both parts of the equation by
$$\frac{1}{3}$$
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
This roots
$$x_{1} = 3 \pi n - 6 + \pi$$
$$x_{2} = 3 \pi n - 2 \pi - 6$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(3 \pi n - 6 + \pi\right) + - \frac{1}{10}$$
=
$$3 \pi n - \frac{61}{10} + \pi$$
substitute to the expression
$$2 \cos{\left(\frac{x}{3} + 2 \right)} \geq 1$$
$$2 \cos{\left(\frac{3 \pi n - \frac{61}{10} + \pi}{3} + 2 \right)} \geq 1$$
/ 1 pi \
2*cos|- -- + -- + pi*n| >= 1
\ 30 3 / but
/ 1 pi \
2*cos|- -- + -- + pi*n| < 1
\ 30 3 / Then
$$x \leq 3 \pi n - 6 + \pi$$
no execute
one of the solutions of our inequality is:
$$x \geq 3 \pi n - 6 + \pi \wedge x \leq 3 \pi n - 2 \pi - 6$$
_____
/ \
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x1 x2
Solving inequality on a graph
Rapid solution
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/ / ___ / 2 \\ / ___ / 2 \\ \ | | 4*tan(1) \/ 3 *\1 + tan (1)/| | 4*tan(1) \/ 3 *\1 + tan (1)/| | And|x <= - 6*atan|----------- - -------------------| + 6*pi, - 6*atan|----------- + -------------------| + 6*pi <= x| | | 2 2 | | 2 2 | | \ \3 - tan (1) 3 - tan (1) / \3 - tan (1) 3 - tan (1) / /
$$x \leq - 6 \operatorname{atan}{\left(- \frac{\sqrt{3} \left(1 + \tan^{2}{\left(1 \right)}\right)}{3 - \tan^{2}{\left(1 \right)}} + \frac{4 \tan{\left(1 \right)}}{3 - \tan^{2}{\left(1 \right)}} \right)} + 6 \pi \wedge - 6 \operatorname{atan}{\left(\frac{\sqrt{3} \left(1 + \tan^{2}{\left(1 \right)}\right)}{3 - \tan^{2}{\left(1 \right)}} + \frac{4 \tan{\left(1 \right)}}{3 - \tan^{2}{\left(1 \right)}} \right)} + 6 \pi \leq x$$
(-6*atan(4*tan(1)/(3 - tan(1)^2) + sqrt(3)*(1 + tan(1)^2)/(3 - tan(1)^2)) + 6*pi <= x)∧(x <= -6*atan(4*tan(1)/(3 - tan(1)^2) - sqrt(3)*(1 + tan(1)^2)/(3 - tan(1)^2)) + 6*pi)
Rapid solution 2
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/ ___ / 2 \\ / ___ / 2 \\
| 4*tan(1) \/ 3 *\1 + tan (1)/| | 4*tan(1) \/ 3 *\1 + tan (1)/|
[- 6*atan|----------- + -------------------| + 6*pi, - 6*atan|----------- - -------------------| + 6*pi]
| 2 2 | | 2 2 |
\3 - tan (1) 3 - tan (1) / \3 - tan (1) 3 - tan (1) /
$$x\ in\ \left[- 6 \operatorname{atan}{\left(\frac{\sqrt{3} \left(1 + \tan^{2}{\left(1 \right)}\right)}{3 - \tan^{2}{\left(1 \right)}} + \frac{4 \tan{\left(1 \right)}}{3 - \tan^{2}{\left(1 \right)}} \right)} + 6 \pi, - 6 \operatorname{atan}{\left(- \frac{\sqrt{3} \left(1 + \tan^{2}{\left(1 \right)}\right)}{3 - \tan^{2}{\left(1 \right)}} + \frac{4 \tan{\left(1 \right)}}{3 - \tan^{2}{\left(1 \right)}} \right)} + 6 \pi\right]$$
x in Interval(-6*atan(sqrt(3)*(1 + tan(1)^2)/(3 - tan(1)^2) + 4*tan(1)/(3 - tan(1)^2)) + 6*pi, -6*atan(-sqrt(3)*(1 + tan(1)^2)/(3 - tan(1)^2) + 4*tan(1)/(3 - tan(1)^2)) + 6*pi)