Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
BlingBlong
Group Title
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
BlingBlong Group Title
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

This Question is Closed

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

TuringTest Group TitleBest 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

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

BlingBlong Group TitleBest 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

TuringTest Group TitleBest 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

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

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

TuringTest Group TitleBest 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

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
welcome!
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.