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erin512

solve the equation on the interval 0≤x≤2pi tanx=0

  • one year ago
  • one year ago

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  1. genius12
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    Erin have you tried solving this yourself? I encourage you to learn from the solutions I provided and try and solve this yourself first. If you get stuck, then feel free to ask.

    • one year ago
  2. erin512
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    thank you, i'm a bit confused but i'll try solving it myself

    • one year ago
  3. genius12
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    One hint is that tan(x) = sin(x) / cos(x) and the saying tan(x) = 0 is the same as saying sin(x) = 0. @erin512

    • one year ago
  4. erin512
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    thanks again, ill try that could you maybe give me an explanation of just one more to help me understand and ill do the next one on my own? i'd really appreciate it

    • one year ago
  5. genius12
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    Remember what I said. tan(x) = 0 is the same as finding x for sin(x) = 0. This is because tan(x) = sin(x) / cos(x). Since tangent is the quotient of sine and cosine, and we are finding the values of x for where it's 0, we are really finding when the quotient sin(x) / cos(x) is 0. The quotient is 0 when sin(x) is 0 because when the numerator is 0, the whole fraction is 0. Where is sin(x) = 0 ? Look at the sine graph if you need to.

    • one year ago
  6. genius12
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    @erin512

    • one year ago
  7. ton12
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    tanx = 0 ---> x = n pie

    • one year ago
  8. ton12
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    Since the intervel is 0≤x≤2pi n = 0,1,2

    • one year ago
  9. erin512
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    n can equal any number?

    • one year ago
  10. erin512
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    oh nevermind thank you

    • one year ago
  11. ton12
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    Welcome

    • one year ago
  12. attila3453
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    Generally, when tanx = a, x = { arctan(a) + k*pi, when k is whole }

    • one year ago
  13. erin512
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    so x = 1, 2?

    • one year ago
  14. erin512
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    i mean x=0,1,2

    • one year ago
  15. DHASHNI
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    \[0 \le x \le 2\pi \] take tan on all \[\tan 0 \le \tan x \le \tan 2\pi\] => \[0 \le tanx \le 0 \] that means tanx =0 \[\ x=tan^{-1} 0= 0 = 2pi \]

    • one year ago
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