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