Sum of all real solutions

philosopher

New member
Joined
Jun 26, 2020
Messages
4
This problem should be easy but for some reason I keep struggling with it.

When i sum all of the solutions i get 20/4π . But the answer is actually 19/4π.
I'm clearly doing something wrong here, it most likely has to do with the range of the functions.
Here's my work.
IMG_E2827.JPG I would appreciate any help. Thanks.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
8,036
When i sum all of the solutions i get 20/4π . But the answer is actually 19/4π.
I'm clearly doing something wrong here, it most likely has to do with the range of the functions.
There are infinitely many real solutions! Surely there must be some restriction stated, such as all solutions in \(\displaystyle [0,2\pi)\). Please state the entire problem exactly as given to you.

But if my guess is right, then you have included two solutions that are not in that interval; and you didn't do a necessary check on each solution to see if it satisfies the condition for its case.
 

philosopher

New member
Joined
Jun 26, 2020
Messages
4
I somehow forgot to add that interval is (0,3π ). Sorry about that.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
9,979
This problem should be easy but for some reason I keep struggling with it.

When i sum all of the solutions i get 20/4π . But the answer is actually 19/4π.
I'm clearly doing something wrong here, it most likely has to do with the range of the functions.
Here's my work. I would appreciate any help. Thanks.
I agree with the given answer. Look here
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
8,036
I somehow forgot to add that interval is (0,3π ). Sorry about that.
Then \(\displaystyle 19\pi/4\) is correct.

Do you see which one of your first three solutions is wrong, and what you need to use in place of your last two? You have the right ideas, but need to pay attention to the sign of the sine, as well as the interval.
 

philosopher

New member
Joined
Jun 26, 2020
Messages
4
Then \(\displaystyle 19\pi/4\) is correct.

Do you see which one of your first three solutions is wrong, and what you need to use in place of your last two? You have the right ideas, but need to pay attention to the sign of the sine, as well as the interval.
Yep, got it now. Thanks a lot!
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
7,374
You just can't say that |sin(x)| = sin (x) or |sin(x)| = -sin (x). This is NOT true.

Rather |sin(x)| = sin (x) if sin(x)>=0 OR |sin(x)| = -sin(x) if sin(x)<0
 
Top