Mister Exam
cos(2*x-3*pi/8)<0 inequation
The solution
You have entered
[src]
/ 3*pi\ cos|2*x - ----| < 0 \ 8 /
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} < 0$$
cos(2*x - 3*pi/8) < 0
Detail solution
Given the inequality:
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
Solve:
Given the equation
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
- this is the simplest trigonometric equation
We get:
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
This equation is transformed to
$$2 x + \frac{\pi}{8} = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$2 x + \frac{\pi}{8} = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$2 x + \frac{\pi}{8} = 2 \pi n$$
$$2 x + \frac{\pi}{8} = 2 \pi n + \pi$$
, where n - is a integer
Move
$$\frac{\pi}{8}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = 2 \pi n - \frac{\pi}{8}$$
$$2 x = 2 \pi n + \frac{7 \pi}{8}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{16}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{16} - \frac{1}{10}$$
substitute to the expression
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} < 0$$
$$\cos{\left(2 \left(\pi n - \frac{\pi}{16} - \frac{1}{10}\right) - \frac{3 \pi}{8} \right)} < 0$$
one of the solutions of our inequality is:
$$x < \pi n - \frac{\pi}{16}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n - \frac{\pi}{16}$$
$$x > \pi n + \frac{7 \pi}{16}$$
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
Solve:
Given the equation
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
- this is the simplest trigonometric equation
with the change of sign in 0
We get:
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} = 0$$
This equation is transformed to
$$2 x + \frac{\pi}{8} = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$2 x + \frac{\pi}{8} = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$2 x + \frac{\pi}{8} = 2 \pi n$$
$$2 x + \frac{\pi}{8} = 2 \pi n + \pi$$
, where n - is a integer
Move
$$\frac{\pi}{8}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = 2 \pi n - \frac{\pi}{8}$$
$$2 x = 2 \pi n + \frac{7 \pi}{8}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{16}$$
$$x_{2} = \pi n + \frac{7 \pi}{16}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{16}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{16} - \frac{1}{10}$$
substitute to the expression
$$\cos{\left(2 x - \frac{3 \pi}{8} \right)} < 0$$
$$\cos{\left(2 \left(\pi n - \frac{\pi}{16} - \frac{1}{10}\right) - \frac{3 \pi}{8} \right)} < 0$$
sin(-1/5 + 2*pi*n) < 0
one of the solutions of our inequality is:
$$x < \pi n - \frac{\pi}{16}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n - \frac{\pi}{16}$$
$$x > \pi n + \frac{7 \pi}{16}$$
Solving inequality on a graph
Rapid solution
[src]
/ / / / /pi\\\ \ / / /pi\\ \ \ | | | |sin|--||| / _____________________\| | |cos|--|| / _____________________\| | | | | | \16/|| | / 2/pi\ 2/pi\ || | | \16/| | / 2/pi\ 2/pi\ || | And|x < -I*|I*|pi - atan|-------|| + log| / cos |--| + sin |--| ||, -I*|I*atan|-------| + log| / cos |--| + sin |--| || < x| | | | | /pi\|| \\/ \16/ \16/ /| | | /pi\| \\/ \16/ \16/ /| | | | | |cos|--||| | | |sin|--|| | | \ \ \ \ \16/// / \ \ \16// / /
$$x < - i \left(\log{\left(\sqrt{\sin^{2}{\left(\frac{\pi}{16} \right)} + \cos^{2}{\left(\frac{\pi}{16} \right)}} \right)} + i \left(\pi - \operatorname{atan}{\left(\frac{\sin{\left(\frac{\pi}{16} \right)}}{\cos{\left(\frac{\pi}{16} \right)}} \right)}\right)\right) \wedge - i \left(\log{\left(\sqrt{\sin^{2}{\left(\frac{\pi}{16} \right)} + \cos^{2}{\left(\frac{\pi}{16} \right)}} \right)} + i \operatorname{atan}{\left(\frac{\cos{\left(\frac{\pi}{16} \right)}}{\sin{\left(\frac{\pi}{16} \right)}} \right)}\right) < x$$
(-i*(i*atan(cos(pi/16)/sin(pi/16)) + log(sqrt(cos(pi/16)^2 + sin(pi/16)^2))) < x)∧(x < -i*(i*(pi - atan(sin(pi/16)/cos(pi/16))) + log(sqrt(cos(pi/16)^2 + sin(pi/16)^2))))