Thread: 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

i dont have a clue about this does sum 1 know???

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

--->

Is it ok ? oO

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

--->

Is it ok ? oO

Last one should be

Right? Probably just a typo.

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

cos5x=0 waaaaaaaaaaaaah

and avoid using chat language such as:

does sum 1 know

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