khaliif6484
khaliif6484 khaliif6484
  • 11-09-2016
  • Mathematics
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(sin(lnx)+cos(lnx)=0

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Panoyin
Panoyin Panoyin
  • 11-09-2016
sin[ln(x)] + cos[ln(x)] = 0
          cos[ln(x)]     cos[ln(x)]
             tan[ln(x)] + 1 = 0
                             - 1  - 1
                  tan[ln(x)] = -1
         tan⁻¹[tan[ln(x)]] = tan⁻(-1)
                         ln(x) = -45
                         e¹ⁿ⁽ˣ⁾ = e⁻⁴⁵
                              x ≈ 2.8 × 10⁻²⁰
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