Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
-
How to use it?
- Inequation:
- (1/4)^2>=8
- (x^2+4)^(1/2)>=-2
- 9^x<=-x+1
- (-x/8)+1/3>0
- Derivative of:
-
cos(2*x)
- Graphing y =:
- cos(2*x)
- Integral of d{x}:
- cos(2*x)
-
Identical expressions
- cos(two *x)>-(√ three)/ two
- co sinus of e of (2 multiply by x) greater than minus (√3) divide by 2
- co sinus of e of (two multiply by x) greater than minus (√ three) divide by two
- cos(2x)>-(√3)/2
- cos2x>-√3/2
- cos(2*x)>-(√3) divide by 2
-
Similar expressions
- cos(2*x)>+(√3)/2
- 1-cos^2x-(1+sqr3)*sinx+sqr3<=0
- sin(x/4)<=-√3/2
- cos^2x-sin^2x>√2/2
- cos(п/3+1/2x)>-√3/2
- sin(x)>cos2*(x)
- cosx≤-√3/2
- cos(2*x+pi/6)<-1/2
- cos(п/3+1/2x)-≥-√3/2
cos(2*x)>-(√3)/2 inequation
The solution
You have entered
[src]
___
-\/ 3
cos(2*x) > -------
2
$$\cos{\left(2 x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
cos(2*x) > (-sqrt(3))/2
Detail solution
Given the inequality:
$$\cos{\left(2 x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$2 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
Or
$$2 x = \pi n + \frac{5 \pi}{6}$$
$$2 x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{5 \pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{5 \pi}{12}$$
substitute to the expression
$$\cos{\left(2 x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{5 \pi}{12}\right) \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
one of the solutions of our inequality is:
$$x < \frac{\pi n}{2} + \frac{5 \pi}{12}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x > \frac{\pi n}{2} - \frac{\pi}{12}$$
$$\cos{\left(2 x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$2 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
Or
$$2 x = \pi n + \frac{5 \pi}{6}$$
$$2 x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{5 \pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{5 \pi}{12}$$
substitute to the expression
$$\cos{\left(2 x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{5 \pi}{12}\right) \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
___
/ 1 pi \ -\/ 3
-sin|- - + -- + pi*n| > -------
\ 5 3 / 2
one of the solutions of our inequality is:
$$x < \frac{\pi n}{2} + \frac{5 \pi}{12}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{2} + \frac{5 \pi}{12}$$
$$x > \frac{\pi n}{2} - \frac{\pi}{12}$$
Solving inequality on a graph
Rapid solution
[src]
/ / / ___ ___\\ / / ___ ___\ \\ | | |\/ 2 + \/ 6 || | |\/ 2 + \/ 6 | || Or|And|0 <= x, x < -atan|-------------||, And|x <= pi, pi + atan|-------------| < x|| | | | ___ ___|| | | ___ ___| || \ \ \\/ 2 - \/ 6 // \ \\/ 2 - \/ 6 / //
$$\left(0 \leq x \wedge x < - \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)}\right) \vee \left(x \leq \pi \wedge \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)} + \pi < x\right)$$
((0 <= x)∧(x < -atan((sqrt(2) + sqrt(6))/(sqrt(2) - sqrt(6)))))∨((x <= pi)∧(pi + atan((sqrt(2) + sqrt(6))/(sqrt(2) - sqrt(6))) < x))
Rapid solution 2
[src]
/ ___ ___\ / ___ ___\
|\/ 2 + \/ 6 | |\/ 2 + \/ 6 |
[0, -atan|-------------|) U (pi + atan|-------------|, pi]
| ___ ___| | ___ ___|
\\/ 2 - \/ 6 / \\/ 2 - \/ 6 /
$$x\ in\ \left[0, - \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)}\right) \cup \left(\operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)} + \pi, \pi\right]$$
x in Union(Interval.Ropen(0, -atan((sqrt(2) + sqrt(6))/(-sqrt(6) + sqrt(2)))), Interval.Lopen(atan((sqrt(2) + sqrt(6))/(-sqrt(6) + sqrt(2))) + pi, pi))