anonymous
  • anonymous
Find all solutions in the interval [0, 2π). 7 tan3x - 21 tan x = 0
Mathematics
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anonymous
  • anonymous
Find all solutions in the interval [0, 2π). 7 tan3x - 21 tan x = 0
Mathematics
chestercat
  • chestercat
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zepdrix
  • zepdrix
Is that \(\large\rm \tan(3x)\) or \(\large\rm \tan^3x\) ? :)
anonymous
  • anonymous
tan ^3 x
zepdrix
  • zepdrix
\[\large\rm 7\tan^3x-21\tan x=0\]Let's factor some stuff first. Looks like they both have..... a tangent... and a 7, ya?

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anonymous
  • anonymous
7tanx(tan^2x-3)=0
zepdrix
  • zepdrix
Ok good! Now we apply our `Zero-Factor Property`, setting each individual factor equal to zero, and then solving from there. So our first factor equal to zero is: \(\large\rm 7\tan x=0\)
zepdrix
  • zepdrix
You can divide by 7 to simplify things down a bit: \(\large\rm \tan x=0\) Understand how to solve for x in this case? :)
anonymous
  • anonymous
would you plug it into a calculator using tan^-1?
zepdrix
  • zepdrix
I guess you `could` do that :p it's better just remember some of your special angles. tangent is 0 when the angle x is 0.
zepdrix
  • zepdrix
Oh but I guess umm... they want all of the solutions from 0 to 2pi, so that produces another angle as well.
anonymous
  • anonymous
oh yeahhhh, so then pi and 0? or would it be 1/pi and 3/pi?
zepdrix
  • zepdrix
sorry got distracted :) um um... yaaaaa, 0 and pi sound good for our first two solutions.
zepdrix
  • zepdrix
Applying the zero-factor property again:\[\large\rm \tan^2x-3=0\]
zepdrix
  • zepdrix
So ummm... how bout.. add 3. then square root or something, ya?
zepdrix
  • zepdrix
\[\large\rm \tan^2=3\]
zepdrix
  • zepdrix
\[\large\rm \tan x=\pm\sqrt{3}\]
zepdrix
  • zepdrix
The positive root 3 will give you 2 angles again. while the negative root 3 will give you 2 more angles! So this solution set is actually giving us 4 angles! kinda crazy.
anonymous
  • anonymous
uhhhhh... how do you get your solutions from \[\pm \sqrt{3}\]
zepdrix
  • zepdrix
Hmm you gotta get better with your unit circle missy! :O http://4.bp.blogspot.com/-FMVojhlkcSQ/UInlIKO88oI/AAAAAAAAAC4/-fzMOz6di2Y/s1600/image010.jpg So umm.. i couldn't find a really good picture, but here is one.
zepdrix
  • zepdrix
The coordinate pair lists first the sine value of that angle, and then the cosine. And then outside of the brackets, on the right, is the tangent value of that angle. So if we look in the first quadrant, it looks like the angle pi/3 gives us sqrt(3), ya?
anonymous
  • anonymous
Ohhhhh! I see now! :P Thank you so much!
zepdrix
  • zepdrix
cool c:

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