Mister Exam
2cosx-1≥0 inequation
The solution
Detail solution
Given the inequality:
$$2 \cos{\left(x \right)} - 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(x \right)} - 1 = 0$$
Solve:
Given the equation
$$2 \cos{\left(x \right)} - 1 = 0$$
- this is the simplest trigonometric equation
We get:
$$2 \cos{\left(x \right)} = 1$$
The equation is transformed to
$$\cos{\left(x \right)} = \frac{1}{2}$$
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$x = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$x = \pi n + \frac{\pi}{3}$$
$$x = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$2 \cos{\left(x \right)} - 1 \geq 0$$
$$2 \cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{3} \right)} - 1 \geq 0$$
but
Then
$$x \leq \pi n + \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq \pi n + \frac{\pi}{3} \wedge x \leq \pi n - \frac{2 \pi}{3}$$
$$2 \cos{\left(x \right)} - 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(x \right)} - 1 = 0$$
Solve:
Given the equation
$$2 \cos{\left(x \right)} - 1 = 0$$
- this is the simplest trigonometric equation
Move -1 to right part of the equation
with the change of sign in -1
We get:
$$2 \cos{\left(x \right)} = 1$$
Divide both parts of the equation by 2
The equation is transformed to
$$\cos{\left(x \right)} = \frac{1}{2}$$
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$x = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$x = \pi n + \frac{\pi}{3}$$
$$x = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n - \frac{2 \pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$2 \cos{\left(x \right)} - 1 \geq 0$$
$$2 \cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{3} \right)} - 1 \geq 0$$
/ 1 pi \
-1 + 2*cos|- -- + -- + pi*n| >= 0
\ 10 3 / but
/ 1 pi \
-1 + 2*cos|- -- + -- + pi*n| < 0
\ 10 3 / Then
$$x \leq \pi n + \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq \pi n + \frac{\pi}{3} \wedge x \leq \pi n - \frac{2 \pi}{3}$$
_____
/ \
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x1 x2
Solving inequality on a graph
Rapid solution
[src]
/ / pi\ /5*pi \\ Or|And|0 <= x, x <= --|, And|---- <= x, x <= 2*pi|| \ \ 3 / \ 3 //
$$\left(0 \leq x \wedge x \leq \frac{\pi}{3}\right) \vee \left(\frac{5 \pi}{3} \leq x \wedge x \leq 2 \pi\right)$$
((0 <= x)∧(x <= pi/3))∨((5*pi/3 <= x)∧(x <= 2*pi))
Rapid solution 2
[src]
pi 5*pi
[0, --] U [----, 2*pi]
3 3
$$x\ in\ \left[0, \frac{\pi}{3}\right] \cup \left[\frac{5 \pi}{3}, 2 \pi\right]$$
x in Union(Interval(0, pi/3), Interval(5*pi/3, 2*pi))