Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
-
How to use it?
- Inequation:
- 0,55^2x−6≥0,55^x+4
- 1/16^9-x<1/3
- 1,5^((x^2+x-20)/x)<=1,5^0
- 2*(x+2)+10>=-11-3*x
- Integral of d{x}:
-
cost
-
1/2
- Derivative of:
-
cost
-
1/2
- Sum of series:
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1/2
-
Identical expressions
- cost> one / two
- co sinus of e of t greater than 1 divide by 2
- co sinus of e of t greater than one divide by two
- cost>1 divide by 2
-
Similar expressions
- (cos(t))^4+(sin(t))^4
- (2x+1+sqrt(4x-11))/2>=6
cost>1/2 inequation
The solution
Detail solution
Given the inequality:
$$\cos{\left(t \right)} > \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(t \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(t \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$t = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$t = \pi n + \frac{\pi}{3}$$
$$t = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
This roots
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$\cos{\left(t \right)} > \frac{1}{2}$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{3} \right)} > \frac{1}{2}$$
Then
$$t < \pi n + \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$t > \pi n + \frac{\pi}{3} \wedge t < \pi n - \frac{2 \pi}{3}$$
$$\cos{\left(t \right)} > \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(t \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(t \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$t = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$t = \pi n + \frac{\pi}{3}$$
$$t = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
This roots
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n - \frac{2 \pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$\cos{\left(t \right)} > \frac{1}{2}$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{3} \right)} > \frac{1}{2}$$
/ 1 pi \ cos|- -- + -- + pi*n| > 1/2 \ 10 3 /
Then
$$t < \pi n + \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$t > \pi n + \frac{\pi}{3} \wedge t < \pi n - \frac{2 \pi}{3}$$
_____
/ \
-------ο-------ο-------
t1 t2
Solving inequality on a graph
Rapid solution 2
[src]
pi 5*pi
[0, --) U (----, 2*pi]
3 3
$$t\ in\ \left[0, \frac{\pi}{3}\right) \cup \left(\frac{5 \pi}{3}, 2 \pi\right]$$
t in Union(Interval.Ropen(0, pi/3), Interval.Lopen(5*pi/3, 2*pi))
Rapid solution
[src]
/ / pi\ / 5*pi \\ Or|And|0 <= t, t < --|, And|t <= 2*pi, ---- < t|| \ \ 3 / \ 3 //
$$\left(0 \leq t \wedge t < \frac{\pi}{3}\right) \vee \left(t \leq 2 \pi \wedge \frac{5 \pi}{3} < t\right)$$
((0 <= t)∧(t < pi/3))∨((t <= 2*pi)∧(5*pi/3 < t))