For now, there's no easy way to get it. You can get at least some of the solutions by adding 2*pi*n to the argument of the trig function:
In [5]: solve(sin(x + 2*pi*n), x) Out[5]: [-2⋅π⋅n] I don't know an easy way to get pi*n, though (in the general case). This trick also works for more complicated functions: In [19]: solve(sin(x + 2*pi*n) - cos(x + 2*pi*n), x) Out[19]: ⎡π⋅(-8⋅n - 3) π⋅(-8⋅n + 1)⎤ ⎢────────────, ────────────⎥ ⎣ 4 4 ⎦ Aaron Meurer On Thu, Mar 8, 2012 at 4:37 AM, dennis <[email protected]> wrote: > Hi, > > i am trying to find the roots of trigonometric functions such as sin(x), > cos(x), ... > > x = Symbol('x') > print( solve(sin(x), x) ) >> [0] > > Is there a way to get all roots of sin(x)? For example something like this: >> [pi*n] > or >> [0, pi], periodicity 2*pi > > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/sympy/-/wVRK0SxmPiYJ. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
