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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?
 2 years ago
 2 years ago
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?
 2 years ago
 2 years ago

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TuringTestBest ResponseYou've already chosen the best response.1
\[2\sin x(\cos x1)\neq0\]try it on your calculator
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
the 2 makes no difference so the assertion is that\[\sin x(\cos x1)=0\]which is just wrong
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
...as an identity I mean
 2 years ago

BlingBlongBest ResponseYou've already chosen the best response.0
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 years ago

TuringTestBest ResponseYou've already chosen the best response.1
\[2\sin x(\cos x1)=0\to\sin x(\cos x1)=0\]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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
but that's just going on that part of the formula
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
ok I see your problem...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\sin(2x)=2\sin x\]\[2\sin x\cos x=2\sin x\]\[\sin x\cos x=\sin x\]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.\[\sin x\cos x\sin x=0\]\[\sin x(\cos x1)=0\]so we have to solve\[\sin x=0\]\[\cos x1=0\]
 2 years ago

BlingBlongBest ResponseYou've already chosen the best response.0
sorry I forgot to mention that the domain is restricted to [0,2pi]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
no big deal with the domain, makes it simpler
 2 years ago

BlingBlongBest ResponseYou've already chosen the best response.0
oh ok I see :) thanks ugh
 2 years ago
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