## BlingBlong 3 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?

1. TuringTest

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

2. TuringTest

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

3. TuringTest

...as an identity I mean

4. BlingBlong

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

5. TuringTest

$2\sin x(\cos x-1)=0\to\sin x(\cos x-1)=0$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

6. TuringTest

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

7. TuringTest

8. TuringTest

$\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$

9. BlingBlong

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

10. TuringTest

no big deal with the domain, makes it simpler

11. BlingBlong

oh ok I see :) thanks ugh

12. TuringTest

welcome!