Function - sin x - ln x = 0, where x is in radians

by **ik4u2** » Fri Apr 15, 2011 8:24 pm

Consider the equation
sin x - ln x = 0, where x is in radians
find an integer 'a' such that the root between 2 and 3 lies between 2 + a/10 and 2 + (a+1)/10

by **graphically** » Sat Apr 16, 2011 2:35 pm

If we solve [tex]sin x - ln x = 0[/tex] graphically:
[tex]x \approx 2.219107148913746[/tex]

by **sin x - ln x** » Sat Apr 16, 2011 2:37 pm

so a = 2
because 2 + 2/10 < 2.219107148913746 < 2 + 3/10

Function - sin x - ln x = 0, where x is in radians

Function - sin x - ln x = 0, where x is in radians

Re: Function - sin x - ln x = 0, where x is in radians

Re: Function - sin x - ln x = 0, where x is in radians

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