equation

[man111]

find all real solution of $\displaystyle x^2-3\lfloor \sin \left(x-\frac{\pi}{6}\right)\rfloor = 3$

where $\lfloor x \rfloor = $ floor function

[Ktina]

man111 в сообщении #586291 писал(а):

find all real solution of $\displaystyle x^2-3\lfloor \sin \left(x-\frac{\pi}{6}\right)\rfloor = 3$

where $\lfloor x \rfloor = $ floor function

The integral part of $\sin (x-\frac{\pi}{6})$ can be only 0, 1 or -1.
So, we need to check exactly 5 options: $x=\pm\sqrt 6,\quad x=\pm\sqrt 3 \quad \text{and}\quad x=0$, it's easy.

-- 18.06.2012, 11:55 --

Where is the problem from?

[TOTAL]

Ktina в сообщении #586310 писал(а):

The integral part

[Ktina]

TOTAL в сообщении #586313 писал(а):

Ktina в сообщении #586310 писал(а):

The integral part

What's wrong with this picture expression?

[TOTAL]

Ktina в сообщении #586315 писал(а):

TOTAL в сообщении #586313 писал(а):

Ktina в сообщении #586310 писал(а):

The integral part

What's wrong with this picture expression?

Integer

[Ktina]

TOTAL в сообщении #586322 писал(а):

Integer

(Оффтоп)

First of all, look here, please. Second, "integer" is a noun and not an adjective. So, the correct form is "integral part" and not "integer part".