Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
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How to use it?
- Inequation:
- 4(x-5)>=3/(x+2)
- (x^2+3^x-3)^5>(x^2+9^x-3^x)*5
- 16^x-9<1
-
cos(x)>=0,2
- Derivative of:
-
cos(x)
- Graphing y =:
-
cos(x)
- Integral of d{x}:
-
cos(x)
-
Identical expressions
- cos(x)>= zero , two
- co sinus of e of (x) greater than or equal to 0,2
- co sinus of e of (x) greater than or equal to zero , two
- cosx>=0,2
- cos(x)>=O,2
-
Similar expressions
- cos(x/5)>(-sqrt(2)/2)
- cosx>=-sqrt3/2
- cosx>=0,2
cos(x)>=0,2 inequation
The solution
Detail solution
Given the inequality:
$$\cos{\left(x \right)} \geq \frac{1}{5}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(x \right)} = \frac{1}{5}$$
Solve:
Given the equation
$$\cos{\left(x \right)} = \frac{1}{5}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
Or
$$x = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
, where n - is a integer
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
This roots
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}\right) - \frac{1}{10}$$
=
$$2 \pi n - \frac{1}{10} + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
substitute to the expression
$$\cos{\left(x \right)} \geq \frac{1}{5}$$
$$\cos{\left(2 \pi n - \frac{1}{10} + \operatorname{acos}{\left(\frac{1}{5} \right)} \right)} \geq \frac{1}{5}$$
one of the solutions of our inequality is:
$$x \leq 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x \geq 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$\cos{\left(x \right)} \geq \frac{1}{5}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(x \right)} = \frac{1}{5}$$
Solve:
Given the equation
$$\cos{\left(x \right)} = \frac{1}{5}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
Or
$$x = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
, where n - is a integer
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
This roots
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x_{2} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}\right) - \frac{1}{10}$$
=
$$2 \pi n - \frac{1}{10} + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
substitute to the expression
$$\cos{\left(x \right)} \geq \frac{1}{5}$$
$$\cos{\left(2 \pi n - \frac{1}{10} + \operatorname{acos}{\left(\frac{1}{5} \right)} \right)} \geq \frac{1}{5}$$
cos(1/10 - acos(1/5)) >= 1/5
one of the solutions of our inequality is:
$$x \leq 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
_____ _____
\ /
-------•-------•-------
x_1 x_2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 2 \pi n + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
$$x \geq 2 \pi n - \pi + \operatorname{acos}{\left(\frac{1}{5} \right)}$$
Solving inequality on a graph
Rapid solution
[src]
/ / / ___\\ / / ___\ \\ Or\And\0 <= x, x <= atan\2*\/ 6 //, And\- atan\2*\/ 6 / + 2*pi <= x, x < 2*pi//
$$\left(0 \leq x \wedge x \leq \operatorname{atan}{\left(2 \sqrt{6} \right)}\right) \vee \left(- \operatorname{atan}{\left(2 \sqrt{6} \right)} + 2 \pi \leq x \wedge x < 2 \pi\right)$$
((0 <= x)∧(x <= atan(2*sqrt(6))))∨((x < 2*pi)∧(-atan(2*sqrt(6)) + 2*pi <= x))
Rapid solution 2
[src]
/ ___\ / ___\ [0, atan\2*\/ 6 /] U [- atan\2*\/ 6 / + 2*pi, 2*pi)
$$x\ in\ \left[0, \operatorname{atan}{\left(2 \sqrt{6} \right)}\right] \cup \left[- \operatorname{atan}{\left(2 \sqrt{6} \right)} + 2 \pi, 2 \pi\right)$$
x in Union(Interval(0, atan(2*sqrt(6))), Interval.Ropen(-atan(2*sqrt(6)) + 2*pi, 2*pi))
The graph