Mister Exam
1+2cosx>0 inequation
The solution
Detail solution
Given the inequality:
$$2 \cos{\left(x \right)} + 1 > 0$$
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
$$x = \pi n + \frac{2 \pi}{3}$$
$$x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{2 \pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{2 \pi}{3}$$
substitute to the expression
$$2 \cos{\left(x \right)} + 1 > 0$$
$$2 \cos{\left(\pi n - \frac{1}{10} + \frac{2 \pi}{3} \right)} + 1 > 0$$
one of the solutions of our inequality is:
$$x < \pi n + \frac{2 \pi}{3}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n + \frac{2 \pi}{3}$$
$$x > \pi n - \frac{\pi}{3}$$
$$2 \cos{\left(x \right)} + 1 > 0$$
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
$$x = \pi n + \frac{2 \pi}{3}$$
$$x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
This roots
$$x_{1} = \pi n + \frac{2 \pi}{3}$$
$$x_{2} = \pi n - \frac{\pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{2 \pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{2 \pi}{3}$$
substitute to the expression
$$2 \cos{\left(x \right)} + 1 > 0$$
$$2 \cos{\left(\pi n - \frac{1}{10} + \frac{2 \pi}{3} \right)} + 1 > 0$$
/ 1 pi \
1 - 2*sin|- -- + -- + pi*n| > 0
\ 10 6 / one of the solutions of our inequality is:
$$x < \pi n + \frac{2 \pi}{3}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n + \frac{2 \pi}{3}$$
$$x > \pi n - \frac{\pi}{3}$$
Solving inequality on a graph
Rapid solution
[src]
/ / 2*pi\ / 4*pi \\ Or|And|0 <= x, x < ----|, And|x <= 2*pi, ---- < x|| \ \ 3 / \ 3 //
$$\left(0 \leq x \wedge x < \frac{2 \pi}{3}\right) \vee \left(x \leq 2 \pi \wedge \frac{4 \pi}{3} < x\right)$$
((0 <= x)∧(x < 2*pi/3))∨((x <= 2*pi)∧(4*pi/3 < x))
Rapid solution 2
[src]
2*pi 4*pi
[0, ----) U (----, 2*pi]
3 3
$$x\ in\ \left[0, \frac{2 \pi}{3}\right) \cup \left(\frac{4 \pi}{3}, 2 \pi\right]$$
x in Union(Interval.Ropen(0, 2*pi/3), Interval.Lopen(4*pi/3, 2*pi))