Question

So I have the function
sin(2x) = 2 sin(x)
=
2sin(x)cos(x) = 2sin(x)
=
(cos(x) − 1) = 0
but the text book claims that
2sin(x) (cos(x) − 1) = 0
Can anyone tell me why this is? Did I overlook something?

Answer 1

\[2\sin x(\cos x-1)\neq0\]try it on your calculator

Answer 2

the 2 makes no difference so the assertion is that\[\sin x(\cos x-1)=0\]which is just wrong

Answer 3

...as an identity I mean

Answer 4

Furthermore it states that the solution to 2sin(x) is equal to
x = 0, π, 2π.
Which doesn't make sense to me seeing as it has the integer 2 in front of it

Answer 5

\[2\sin x(\cos x-1)=0\to\sin x(\cos x-1)=0\]because 2 cannot be zero, so we have\[\sin x=0\to x=n\pi, n\in\mathbb N\]but you may have figure that along with the cosine part

Answer 6

but that's just going on that part of the formula

Answer 7

ok I see your problem...

Answer 8

\[\sin(2x)=2\sin x\]\[2\sin x\cos x=2\sin x\]\[\sin x\cos x=\sin x\]now you can't just divide by sinx because that assumes that sinx is not zero, which it might be!, so we have to keep it around and factor if we want all the solutions.\[\sin x\cos x-\sin x=0\]\[\sin x(\cos x-1)=0\]so we have to solve\[\sin x=0\]\[\cos x-1=0\]

Answer 9

sorry I forgot to mention that the domain is restricted to [0,2pi]

Answer 10

no big deal with the domain, makes it simpler

Answer 11

oh ok I see :) thanks ugh

Answer 12

welcome!