Solve the following equation on the set of integers:
$\displaystyle \sin(\frac{\pi}{3}(x-\sqrt{x^2-3x-12}))=0$
Hello, perash!
I haven't solve it yet . . . It's quite involved . . .
The Captain is absolutely correct . . .Solve the following equation on the set of integers:
$\displaystyle \sin\left[\frac{\pi}{3}\left(x-\sqrt{x^2-3x-12}\right)\right]\:=\:0$
We have: .$\displaystyle \sin\left[\pi\left(\frac{x - \sqrt{x^2-3x-12}}{3}\right)\right] \:=\:0$
We know that: .$\displaystyle \sin(\pi k) \:=\:0$ . . . for any integer $\displaystyle k.$
Hence: .$\displaystyle \frac{x-\sqrt{x^2-3x-12}}{3}$ must be an integer.
. . We can let: .$\displaystyle \frac{x-\sqrt{x^2-3x-12}}{3} \:=\:k$
There are more restrictions:
. . $\displaystyle x^2-3x-12$ must be nonnegative: .$\displaystyle x^2-3x-12 \:\geq\:0\quad\Rightarrow\quad x \leq -3,\;x \geq 6$
. . And, of course, $\displaystyle x^2-3x-12$ must be a square.
Okay, my stab at this is confusing me.
$\displaystyle x - \sqrt{x^2 - 3x - 12} \equiv 0~\text{ mod 3}$
$\displaystyle x - \sqrt{x^2} \equiv 0~\text{ mod 3}$
$\displaystyle x - |x| \equiv 0~\text{ mod 3}$
Isn't this an identity in mod 3? But the original equation is not true for all x within the domain.
-Dan
16 doesn't work.
I have "reduced" the problem to solving the Diophantine equation:
$\displaystyle n^2 - 4m^2 = 57$
(I can explain how I got this, but I'd like the original poster to do some work on their own.)
So
$\displaystyle (n + 2m)(n - 2m) = 57$
There are going to be some obvious limits on m and n here because n + 2m has to be one of 1, 3, 19, 57.
So we are looking the solution to one of two systems of equations:
$\displaystyle n + 2m = 57$
and
$\displaystyle n - 2m = 1$
or
$\displaystyle n + 2m = 19$
and
$\displaystyle n - 2m = 3$
The first has a solution of m = 14 and n = 29, giving an x value of -13. The second has a solution of m = 4 and n = 11, giving an x value of 7.
So the only two solutions are x = -13 and x = 7.
-Dan