n=1+sinx/n.....n=?
回答 (2)
2017-07-28 10:21 pm
Case 1 : n = (1 + sinx) / n
n = (1 + sinx) /n
n² = 1 + sinx
n = √(1 + sinx)
====
Case 2 : n = 1 + (sinx/n)
n = 1 + (sinx/n)
n = (n/n) + (sinx/n)
n = (n + sinx) / n
n² = n + sinx
n² - n - sinx = 0
n = {-(-1) ± √[(-1)² + 4*1*(-sinx)]} / 2
n = [1 + √(1 + 4sinx)] / 2 or n = [1 - √(1 + 4sinx)] / 2
n = (1 + sinx) /n
n² = 1 + sinx
n = √(1 + sinx)
====
Case 2 : n = 1 + (sinx/n)
n = 1 + (sinx/n)
n = (n/n) + (sinx/n)
n = (n + sinx) / n
n² = n + sinx
n² - n - sinx = 0
n = {-(-1) ± √[(-1)² + 4*1*(-sinx)]} / 2
n = [1 + √(1 + 4sinx)] / 2 or n = [1 - √(1 + 4sinx)] / 2
2017-07-28 10:00 pm
question makes no sense
if you want to solve
N=1+sinx/n
for n, then do this
(N-1) = sinx/n
1/(N-1) = n/sinx
sinx/(N-1) = n
if you want to solve
N=1+sinx/n
for n, then do this
(N-1) = sinx/n
1/(N-1) = n/sinx
sinx/(N-1) = n
收錄日期: 2021-04-18 17:32:59
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