Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
-
How to use it?
- Inequation:
- (-2x-9)/(x+2)(x-3)<0
- (x^2+x-20)/(x^2-6*x+y)>0
- (x-5)*(sqrt(9-x))>0
- log(x)/log(3)<11-x
- Graphing y =:
- cos(2*x+pi/6)
- Derivative of:
- -1/2
- Integral of d{x}:
-
-1/2
- Limit of the function:
- -1/2
-
Identical expressions
- cos(two *x+pi/ six)<- one / two
- co sinus of e of (2 multiply by x plus Pi divide by 6) less than minus 1 divide by 2
- co sinus of e of (two multiply by x plus Pi divide by six) less than minus one divide by two
- cos(2x+pi/6)<-1/2
- cos2x+pi/6<-1/2
- cos(2*x+pi divide by 6)<-1 divide by 2
-
Similar expressions
- cos(2x)>-(1/2)
- (1/2)^(x-2)-(1/2)^x<=3
- cos(2*x-pi/6)<-1/2
- x^2-2x-1/(2x-5)(x+2)^2<0
- ((3*2^(x+1)-5)/(2^(x-1)))^2+((3*2^(x+1)-1)/(2^(x-1)-2))^2≤(9*4^(x-0,5)-9*2^(x+1)+10)/(4^x(-1)-2^x)
- cos(2*x+pi/6)<+1/2
cos(2*x+pi/6)<-1/2 inequation
The solution
You have entered
[src]
/ pi\ cos|2*x + --| < -1/2 \ 6 /
$$\cos{\left(2 x + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
cos(2*x + pi/6) < -1/2
Detail solution
Given the inequality:
$$\cos{\left(2 x + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{6} = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$2 x + \frac{\pi}{6} = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$2 x + \frac{\pi}{6} = \pi n + \frac{2 \pi}{3}$$
$$2 x + \frac{\pi}{6} = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = \pi n + \frac{\pi}{2}$$
$$2 x = \pi n - \frac{\pi}{2}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{4}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{4}$$
substitute to the expression
$$\cos{\left(2 x + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{4}\right) + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
but
Then
$$x < \frac{\pi n}{2} + \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{2} + \frac{\pi}{4} \wedge x < \frac{\pi n}{2} - \frac{\pi}{4}$$
$$\cos{\left(2 x + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{6} = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$2 x + \frac{\pi}{6} = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$2 x + \frac{\pi}{6} = \pi n + \frac{2 \pi}{3}$$
$$2 x + \frac{\pi}{6} = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = \pi n + \frac{\pi}{2}$$
$$2 x = \pi n - \frac{\pi}{2}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{4}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{4}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{4}$$
substitute to the expression
$$\cos{\left(2 x + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{4}\right) + \frac{\pi}{6} \right)} < - \frac{1}{2}$$
/ 1 pi \
-sin|- - + -- + pi*n| < -1/2
\ 5 6 / but
/ 1 pi \
-sin|- - + -- + pi*n| > -1/2
\ 5 6 / Then
$$x < \frac{\pi n}{2} + \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{2} + \frac{\pi}{4} \wedge x < \frac{\pi n}{2} - \frac{\pi}{4}$$
_____
/ \
-------ο-------ο-------
x1 x2
Solving inequality on a graph
Rapid solution
[src]
/ / ___ ___\\ |pi |\/ 2 + \/ 6 || And|-- < x, x < pi + atan|-------------|| |4 | ___ ___|| \ \\/ 2 - \/ 6 //
$$\frac{\pi}{4} < x \wedge x < \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)} + \pi$$
(pi/4 < x)∧(x < pi + atan((sqrt(2) + sqrt(6))/(sqrt(2) - sqrt(6))))
Rapid solution 2
[src]
/ ___ ___\
pi |\/ 2 + \/ 6 |
(--, pi + atan|-------------|)
4 | ___ ___|
\\/ 2 - \/ 6 /
$$x\ in\ \left(\frac{\pi}{4}, \operatorname{atan}{\left(\frac{\sqrt{2} + \sqrt{6}}{- \sqrt{6} + \sqrt{2}} \right)} + \pi\right)$$
x in Interval.open(pi/4, atan((sqrt(2) + sqrt(6))/(-sqrt(6) + sqrt(2))) + pi)