correct

Hi smiyc86;

Here is a quick and very ugly way to get the real roots of   (sinx)^4-(sinx)^5-(cosx)^5 = 2

First I notice that big 2 on the rhs and guess that one of the extrema of sin(x) and cos(x) are the answer. We have:

Just trying the first one yields (-1)^4 - (-1)^5 - 0 which equals 2 so x = -π/2 is one solution. Since sin is periodic

with c=0,1,2,3... are all solutions.

There are also complex solutions but I could not get them without resorting to an iterative method so I left them out.

bobbym

Last edited by bobbym (2009-04-17 22:06:33)

Thanks for providing me with another way to do it.

# 3

Find the smallest number N which has the following properties:
its decimal representation has 6 as the last digit.

If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as great as the original number N.

nice ....