# Revision history [back]

This smells of homework. Therefore, some hints .

You're trying to find the zeroes of $\displaystyle-\frac{\sin\left(t\right)}{\cos\left(2 \, t\right)}$ (expression quite easily obtained with Sage). A plot of this function (as easily obtained...) should help to convince you that this function has exactly one zero in the open interval $(0~2\pi)$ of $\mathbb{R}$. Proving it formally is also a good exercice.

However, this plot might also attract your attention to this function's _poles_, which aren't quite the same thing...

This smells of homework. Therefore, some hints .

You're trying to find the zeroes of $\displaystyle-\frac{\sin\left(t\right)}{\cos\left(2 \, t\right)}$ (expression quite easily obtained with Sage). A plot of this function (as easily obtained...) should help to convince you that this function has exactly one zero in the open interval $(0~2\pi)$ of $\mathbb{R}$. Proving it formally is also a good exercice.

However, this plot might also attract your attention to this function's _poles_, poles, which aren't quite the same thing...

Note : This was a good answer to the bad question. See below...

This smells of homework. Therefore, some hints .

You're trying to find the zeroes of $\displaystyle-\frac{\sin\left(t\right)}{\cos\left(2 \, t\right)}$ (expression quite easily obtained with Sage). A plot of this function (as easily obtained...) should help to convince you that this function has exactly one zero in the open interval $(0~2\pi)$ of $\mathbb{R}$. Proving it formally is also a good exercice.

However, this plot might also attract your attention to this function's poles, which aren't quite the same thing...

Now for the _right_ question :

Indeed, the default solver (Maxima's) gives only a partial solution to the real equation (i. e. $\displaystyle\frac{\cos(2t)}{sin(t)}=0$).

One can obtain a better set of solution with:

solve(cos(2*t),t,to_poly_solve="force")
[t == 1/4*pi + 1/2*pi*z40]


(see solve's doc). But this doesn't fulfills the constraint $0\lt t\lt2\pi$.

Note that sympy's solver is not really better :

sage: [s._sage_() for s in sympy.solve(cos(2*t),t)]
[1/4*pi, 3/4*pi]


Mathematica seems to be able to solve the system, but its answer can't currently be parsed back in Sage :

sage: S=[t>0,t<2*pi,cos(2*t)==0]
sage: mathematica.Solve(S,t)
{{t -> Pi/4}, {t -> (3*Pi)/4}, {t -> (5*Pi)/4}, {t -> (7*Pi)/4}}