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(sin(lnx)+cos(lnx)=0

User Xiaoyu Xu
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1 Answer

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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
User Closure
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