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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?

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\[2\sin x(\cos x-1)\neq0\]try it on your calculator
the 2 makes no difference so the assertion is that\[\sin x(\cos x-1)=0\]which is just wrong an identity I mean

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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
\[2\sin x(\cos x-1)=0\to\sin x(\cos x-1)=0\]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
but that's just going on that part of the formula
ok I see your problem...
\[\sin(2x)=2\sin x\]\[2\sin x\cos x=2\sin x\]\[\sin x\cos x=\sin x\]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\[\sin x=0\]\[\cos x-1=0\]
sorry I forgot to mention that the domain is restricted to [0,2pi]
no big deal with the domain, makes it simpler
oh ok I see :) thanks ugh

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