$\displaystyle y=sin6x$
and
$\displaystyle y=sin1/4x$
In other words what might the zeroes be?
As masters wrote in this thread (http://www.mathhelpforum.com/math-he...lp-answer.html) the b is important. For the 1st one, b = 6, and for the 2nd one, b = 1/4.
(BTW, your 2nd one in ambiguous. Some people may think that you mean $\displaystyle y = \sin \frac{1}{4x}$ instead of $\displaystyle y = \sin \frac{1}{4}x$.)
You have to remember the zeros for y = sin x, which is
$\displaystyle 0,\; \pi,\; 2\pi,\; 3\pi,...$;
in other words, multiples of $\displaystyle \pi$.
So to find the zeros of y = sin bx, divide the zeros I listed above by b. For y = sin 6x, the zeros are
$\displaystyle 0,\; \frac{\pi}{6}\; \frac{\pi}{3}\; \frac{\pi}{2},...$.
I'll let you try the other one.
01
Let $\displaystyle u = 6x$ . So now we have $\displaystyle y=\sin{u}$ . For which values of $\displaystyle u$ does $\displaystyle y=\sin{u} = 0$ ?
Now use this and the fact that $\displaystyle x = \frac{u}{6}$ to find out the zeroes to $\displaystyle y=\sin{6x}$
EDIT: tooo slow