No. But 0 is a solution. 0 + cos(0) = 0 + 1 = 1.
Let's analyse if there's another solution. Put f(x) - x + cos(x). Then, f'(x) = 1 - sin(x) >=0 for every x. . We have f'(x) = 0 whenever sin(x) =1, that is, when x = 2kpi + pi/2. But these points are not extermum points, because f' doesn't change from positive to negative, we have f'(x) >=0 for every x. So, f is increasing, so that f vanishes if and only if x = 0. No other solutio,
2007-08-28 03:24:19
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answer #1
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answered by Steiner 7
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it is 1
2007-08-28 10:15:41
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answer #2
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answered by Navita 3
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The value of cos0 is 1. So if we suppose X is 0, then X +cosx=0+cos0=0+1=1
Therefore the value of X is 0 and not 1.
2007-08-28 10:19:46
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answer #3
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answered by spandanaraj b 2
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i think zero, because cos0=1 then according to question
if you put x=0 then
0 + cos 0 = 1
2007-08-28 10:54:22
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answer #4
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answered by Anonymous
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zero.
Cos(0) = 1.
So, 0 + Cos(0) = 1.
2007-08-28 10:16:26
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answer #5
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answered by Anonymous
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i don't know I'm only in grade 8
2007-08-28 10:14:55
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answer #6
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answered by vidkid234 3
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zero
2007-08-31 10:34:07
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answer #7
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answered by Anonymous
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