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I have a problem by using the bisection method.

I have to get a route of 2xcos(2x)-sin(2x)=0 in the interval (3,4)

However by the first estimation, I got a positive number when I put f(3.5) therefore the next interval is (3,3.5). However the solution should be in the direction of 3.8. Does this method just don't work here?

Many thanks in advanced

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    May have solutions in each half of $[3,4].$ Suggest you check the graph... You need subintervals where signs are different at ends. – coffeemath Feb 22 '15 at 21:55
  • Adam: Are you sure you're in radian mode? And I assume it is $2x\cos(2x)-\sin(2x)$ where the $x$ there is not the multiplication sign... With this the sign is + at 3, 3.5 and - at 4. – coffeemath Feb 22 '15 at 22:03
  • I don't get it. It is positive in $3.5$ and negative in $4$, so where is the problem? Just continue with $3.75$.. http://www.wolframalpha.com/input/?i=f%28x%29%3D2xcos%282x%29-sin%282x%29%2C+x+from+3+to+4 – Peter Franek Feb 22 '15 at 22:24

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