Hi,

I need to find

I need to find

**all**roots the equatation has on a specified interval with preset accuracy. The equatation is the following: x^3*tan(x^2-1)+6*x^4=0. Of course, I can find a single root of this equatation (by the way, one of them is 0). But what about others? There are about 25 of them (following to MathCAD), but what about programming way of finding roots of equatation? I know about Newton's, line dividing, iterations way of finding, but they manage to get just 1 root on a interval, that is sure to have just 1 root. What are the methods of solving equatations like this? Note, I am interested in all roots the equatation has with specified accuracy, because some security issuses are depending on them.Hi,

I need to find

**all**roots the equatation has on a specified interval with preset accuracy. The equatation is the following: x^3*tan(x^2-1)+6*x^4=0. Of course, I can find a single root of this equatation (by the way, one of them is 0). But what about others? There are about 25 of them (following to MathCAD), but what about programming way of finding roots of equatation? I know about Newton's, line dividing, iterations way of finding, but they manage to get just 1 root on a interval, that is sure to have just 1 root. What are the methods of solving equatations like this? Note, I am interested in all roots the equatation has with specified accuracy, because some security issuses are depending on them.

Well what is the interval?....if its -2PI to 2PI i can't see why you can't just use the FPU.

The part 6*x^4 is always positive so ....x^3*tan(x^2-1) has to be equal in value but opposite sign to get a root. (Other than the obviouse when x = 0)

So 6*x^4 = - (x^3*tan(x^2-1))

6*x^4 = - (x^3) * 6 * x

Therefore - 6 * x = tan(x^2-1) when the equation is zero or

- x = tan(x^2-1)

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6

However since the FPU only will give a subset of the real numbers you may get some values that are close to zero and may actually be a zero root.

Sorry if im not clear...been a while for math problems.

:alright: :alright: