Almost done! I just have this problem and one more after that. I could really use some guidance with this one. A bit confusing for me. :/ Click attached
(1-cos(x))/sin(x) = sin(x) 1-cos(x) = sin(x)*sin(x) 1-cos(x) = sin^2(x) 1-cos(x) = 1-cos^2(x) 1-z = 1-z^2 ... Let z = cos(x) Now solve for z
It's not matching up, I'm not getting it :/
What solution(s) are you getting for z?
How far did you get if you were unable to solve for z?
really what we have here is the set is answer three check me if I am right?
you there still?
Oh sorry, I left my computer for a bit. Oh, @godorovg how do you come to that conclusion?
(0, pi)
When you solve 1-z = 1-z^2 for z, what do you get?
I went over it again a got option three. Thank you for the help guys!
it's not option 3
Can you please go through it step by step with me then? I can't do this on my own :/
well we know 1-z = 1=z^2 is 1-z = (1-z x 1+z) I am on the right track?
if x = pi, then sin(x) = sin(pi) = 0, which is a no-no because we'll be dividing by zero So x = pi isn't a solution (it's not even in the domain)
1-z = 1-z^2 1-z+z = 1-z^2 + z 1 = 1-z^2+z 1-1 = 1-z^2+z-1 0 = -z^2+z -z^2+z = 0 -z(z-1) = 0 -z = 0 or z-1 = 0 z = 0 or z = 1 -------------------- We made z = cos(x), so cos(x) = 0 or cos(x) = 1 Solve each for x cos(x) = 0 x = arccos(0) x = pi/2 or x = 3pi/2 ----------- cos(x) = 1 x = arccos(1) x = 0 So the solution set is {0, pi/2, 3pi/2}
Oh, I see. Thank you @jim_thompson5910 for the explanation and clarification. I appreciate your help
np
I'm glad it's making sense now
oh wait, I'm not thinking...x = 0 doesn't work (for the same reason I explained above) So it's just {pi/2, 3pi/2}
sry it's late over here lol
x = 0 isn't a solution because it causes a division by zero error
Oh lol it's fine. Thanks!
alright, yw
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