Question

sinx + sinxcosx = 0

Answer 1

Is this correct?

sinx + sinxcosx = 0

sinx (1 + cosx) = 0

sinx = 0 1 + cosx = 0
-1 -1
cosx = -1

(on unit circle for general solutions) x= pi + 2pi n

Answer 2

could you subtract since from both sides then end up with cosx=-1 anyway? instead of sinx=0

Answer 3

Subtract what?

Answer 4

I factored out a sinx

Answer 5

sinx+sinxcosx=0
-sinx -sinx
-------------------
sinxcosx=-sinx
divide by sinx
cosx=-1

Answer 6

Yes I think that would work as well but you get the same answer except by factoring there is an extraneous solution

Answer 7

You have two cases: \(\sin x=0\) and \(\cos x=-1\):

\(\sin x = 0\) is true for \(x=\dfrac{\pi}{2}+\pi n,~n\in\mathbb Z\)


\(\cos x = -1\) is true for \(x=\pi+2\pi n,~n\in\mathbb Z\)

In unit circle, \(\sin x + \sin x \cos x = 0\) at these points:

Answer 8

Is the way I used right?

Answer 9

it's been a while since i've done this so lets wait for someone of greater mind to help us. i'm interested to learn it too.

Answer 10

So I would say that general solution is something like \(x = n\dfrac{\pi}{2},~n\in\mathbb Z, ~n\not|~~4\)

In other word, n can be any integer, except that it is not divide 4.

Answer 11

@geerky42 is my new answer x = pi + 2pi n, 2pi + 2pi n

Answer 12

I believe answer should be \(x = \pi + 2\pi n,~\dfrac{\pi}{2} + \pi n\)