1. ## cos5x=0 waaaaaaaaaaaaah

could someone explain this working,makes no sense to me,Q is solve sin5x=0 in range 0 to 2pie inclusive, giving roots as ,multiples of pi

sin5x=0 implies 5x=kpie , k=0 to 9
implies x=kpie/5 therefore k=0 to9

2. Hello,

sin5x=0 implies 5x=kpie , k=0 to 9
implies x=kpie/5 therefore k=0 to9
I don't understand your writing, though the result is (almost) correct. But the "therefore" part makes no sense...

Let X=5x. Then X is between 0 and 10 pi.
So you have to solve for sin(X)=0

As you said, X=k*pi, with k=..., -2, -1, 0, 1, 2, ....

Since X is between 0 and 10 pi, therefore k=0, 1, ..., 9, 10 (0 and 2 pi are inclusive).

Solutions are $\displaystyle X=0, \ \pi, \ 2 \pi, \dots , \ 9 \pi, \ 10 \pi$

---> $\displaystyle x=0, \frac{\pi}{5}, \frac{2 \pi}{5}, \dots , \frac{9 \pi}{5}, 2 \pi$

Is it ok ? oO

3. Originally Posted by Moo
Hello,

I don't understand your writing, though the result is correct. But this implication makes no sense...

Let X=5x. Then X is between 0 and 10 pi.
So you have to solve for sin(X)=0

As you said, X=k*pi, with k=..., -2, -1, 0, 1, 2, ....

Since X is between 0 and 10 pi, therefore k=0, 1, ..., 9, 10 (0 and 2 pi are inclusive).

Solutions are $\displaystyle X=0, \ \pi, \ 2 \pi, \dots , \ 9 \pi, \ 10 \pi$

---> $\displaystyle x=0, \frac{\pi}{5}, \frac{2 \pi}{5}, \dots , \frac{9 \pi}{5}, 10 \pi$

Is it ok ? oO
Last one should be $\displaystyle \frac{10\pi}{5}$

Right? Probably just a typo.

4. first of all try to give a meaningful title to your post instead of :

and avoid using chat language such as:
does sum 1 know
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$\displaystyle \begin{gathered} \sin 5x = 0 \hfill \\ \Leftrightarrow 5x = \pi k,\quad k \in Z \hfill \\ \Leftrightarrow x = \frac{\pi } {5}k \hfill \\ \end{gathered}$

now we are interested in the answers which are in the following interval [0, 2pi]:

$\displaystyle \begin{gathered} 0 \leqslant \frac{\pi } {5}k \leqslant 2\pi \hfill \\ \hfill \\ \leftrightarrow 0 \leqslant k \leqslant 10 \hfill \\ \end{gathered}$