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anonymous

  • one year ago

Find all solutions in the interval [0, 2π). 7 tan3x - 21 tan x = 0

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  1. zepdrix
    • one year ago
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    Is that \(\large\rm \tan(3x)\) or \(\large\rm \tan^3x\) ? :)

  2. anonymous
    • one year ago
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    tan ^3 x

  3. zepdrix
    • one year ago
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    \[\large\rm 7\tan^3x-21\tan x=0\]Let's factor some stuff first. Looks like they both have..... a tangent... and a 7, ya?

  4. anonymous
    • one year ago
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    7tanx(tan^2x-3)=0

  5. zepdrix
    • one year ago
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    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: \(\large\rm 7\tan x=0\)

  6. zepdrix
    • one year ago
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    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? :)

  7. anonymous
    • one year ago
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    would you plug it into a calculator using tan^-1?

  8. zepdrix
    • one year ago
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    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.

  9. zepdrix
    • one year ago
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    Oh but I guess umm... they want all of the solutions from 0 to 2pi, so that produces another angle as well.

  10. anonymous
    • one year ago
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    oh yeahhhh, so then pi and 0? or would it be 1/pi and 3/pi?

  11. zepdrix
    • one year ago
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    sorry got distracted :) um um... yaaaaa, 0 and pi sound good for our first two solutions.

  12. zepdrix
    • one year ago
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    Applying the zero-factor property again:\[\large\rm \tan^2x-3=0\]

  13. zepdrix
    • one year ago
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    So ummm... how bout.. add 3. then square root or something, ya?

  14. zepdrix
    • one year ago
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    \[\large\rm \tan^2=3\]

  15. zepdrix
    • one year ago
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    \[\large\rm \tan x=\pm\sqrt{3}\]

  16. zepdrix
    • one year ago
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    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.

  17. anonymous
    • one year ago
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    uhhhhh... how do you get your solutions from \[\pm \sqrt{3}\]

  18. zepdrix
    • one year ago
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    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.

  19. zepdrix
    • one year ago
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    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?

  20. anonymous
    • one year ago
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    Ohhhhh! I see now! :P Thank you so much!

  21. zepdrix
    • one year ago
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    cool c:

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