Mister Exam
cos6x≥-1/2 inequation
The solution
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cos(6*x) >= -1/2
$$\cos{\left(6 x \right)} \geq - \frac{1}{2}$$
cos(6*x) >= -1/2
Detail solution
Given the inequality:
$$\cos{\left(6 x \right)} \geq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$6 x = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$6 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$6 x = \pi n + \frac{2 \pi}{3}$$
$$6 x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
This roots
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{6} + \frac{\pi}{9}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}$$
substitute to the expression
$$\cos{\left(6 x \right)} \geq - \frac{1}{2}$$
$$\cos{\left(6 \left(\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}\right) \right)} \geq - \frac{1}{2}$$
one of the solutions of our inequality is:
$$x \leq \frac{\pi n}{6} + \frac{\pi}{9}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x \geq \frac{\pi n}{6} - \frac{\pi}{18}$$
$$\cos{\left(6 x \right)} \geq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$6 x = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$6 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$6 x = \pi n + \frac{2 \pi}{3}$$
$$6 x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
This roots
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{6} + \frac{\pi}{9}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}$$
substitute to the expression
$$\cos{\left(6 x \right)} \geq - \frac{1}{2}$$
$$\cos{\left(6 \left(\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}\right) \right)} \geq - \frac{1}{2}$$
/ 3 pi \
-sin|- - + -- + pi*n| >= -1/2
\ 5 6 / one of the solutions of our inequality is:
$$x \leq \frac{\pi n}{6} + \frac{\pi}{9}$$
_____ _____
\ /
-------•-------•-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x \geq \frac{\pi n}{6} - \frac{\pi}{18}$$
Solving inequality on a graph
Rapid solution
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$$\left(0 \leq x \wedge x \leq - i \left(\log{\left(\sqrt{\sin^{2}{\left(\frac{\pi}{9} \right)} + \cos^{2}{\left(\frac{\pi}{9} \right)}} \right)} + i \operatorname{atan}{\left(\frac{\sin{\left(\frac{\pi}{9} \right)}}{\cos{\left(\frac{\pi}{9} \right)}} \right)}\right)\right) \vee \left(x \leq \frac{\pi}{3} \wedge - i \left(\log{\left(\sqrt{\sin^{2}{\left(\frac{2 \pi}{9} \right)} + \cos^{2}{\left(\frac{2 \pi}{9} \right)}} \right)} + i \operatorname{atan}{\left(\frac{\sin{\left(\frac{2 \pi}{9} \right)}}{\cos{\left(\frac{2 \pi}{9} \right)}} \right)}\right) \leq x\right)$$
((0 <= x)∧(x <= -i*(i*atan(sin(pi/9)/cos(pi/9)) + log(sqrt(cos(pi/9)^2 + sin(pi/9)^2)))))∨((x <= pi/3)∧(-i*(i*atan(sin(2*pi/9)/cos(2*pi/9)) + log(sqrt(cos(2*pi/9)^2 + sin(2*pi/9)^2))) <= x))