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Sympy offers us a solution, but sage is not (yet) ready to parse it totally. But with a little Sympy's doc reading and a little manual unscrewing, one can get it:

sage: [u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
[2*pi*n + arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 pi + 2*pi*n - arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 -pi + 2*pi*n + arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3))),
 2*pi + 2*pi*n - arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3)))]

A bit of fooling around with latex and string splicing allows us to get the $\LaTeX$ form:

$$2 \pi n + \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$\pi + 2 \pi n - \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$-\pi + 2 \pi n + \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$ $$2 \pi + 2 \pi n - \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$

Note: Maxima could also get it (via to_poly_solve), but gives pompous pedantic nonsense involving logs of complex expressions, probably due to the domain:complex default...

Note 2: Substituting these values in c_p gets rid of the trig functions easily (via trig_reduce()). However, the resulting expressions involve sums ad products of scalar and square roots values ; proving them to be 0 is another possibly nontrivial problem... It can be "checked" numerically, and formally proved for the last two:

sage: L2=[u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
sage: [c_p.subs(theta==u).trig_reduce().n() for u in L2]
[2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16]

sage: [bool(c_p.subs(theta==u).trig_reduce()==0) for u in L2]
[False, False, True, True]

Sympy offers us a solution, but sage is not (yet) ready to parse it totally. But with a little Sympy's doc reading and a little manual unscrewing, one can get it:

sage: [u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
[2*pi*n + arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 pi + 2*pi*n - arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 -pi + 2*pi*n + arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3))),
 2*pi + 2*pi*n - arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3)))]

A bit of fooling around with latex and string splicing allows us to get the $\LaTeX$ form:

$$2 \pi n + \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$\pi + 2 \pi n - \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$-\pi + 2 \pi n + \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$ $$2 \pi + 2 \pi n - \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$

Note: Maxima could also get it (via to_poly_solve), but gives pompous pedantic nonsense involving logs of complex expressions, probably due to the domain:complex default...

Note 2: Substituting these values in c_p gets rid of the trig functions easily (via trig_reduce()). after declaring n integer...). However, the resulting expressions involve sums ad products of scalar and square roots values ; proving them to be 0 is another possibly nontrivial problem... It can be "checked" numerically, and formally proved for the last two:

sage: L2=[u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
sage: [c_p.subs(theta==u).trig_reduce().n() for u in L2]
[2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16]

sage: [bool(c_p.subs(theta==u).trig_reduce()==0) for u in L2]
[False, False, True, True]

Sympy offers us a solution, but sage is not (yet) ready to parse it totally. But with a little Sympy's doc reading and a little manual unscrewing, one can get it:

sage: [u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
[2*pi*n + arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 pi + 2*pi*n - arctan((2*pi - 5)/(pi*sqrt(20/pi - 25/pi^2 + 12))),
 -pi + 2*pi*n + arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3))),
 2*pi + 2*pi*n - arctan(1/2*(2*pi + 5)/(pi*sqrt(-5/pi - 25/4/pi^2 + 3)))]

A bit of fooling around with latex and string splicing allows us to get the $\LaTeX$ form:

$$2 \pi n + \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$\pi + 2 \pi n - \arctan\left(\frac{2 \pi - 5}{\pi \sqrt{\frac{20}{\pi} - \frac{25}{\pi^{2}} + 12}}\right)$$ $$-\pi + 2 \pi n + \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$ $$2 \pi + 2 \pi n - \arctan\left(\frac{2 \pi + 5}{2 \pi \sqrt{-\frac{5}{\pi} - \frac{25}{4 \pi^{2}} + 3}}\right)$$

Note: Maxima could also get it (via to_poly_solve), but gives pompous pedantic nonsense involving logs of complex expressions, probably due to the domain:complex default...

Note 2: Substituting these values in c_p gets rid of the trig functions easily (via trig_reduce() after declaring n integer...). However, the resulting expressions involve sums ad products of scalar and square roots values ; proving them to be 0 is another possibly nontrivial problem... It can be "checked" numerically, and formally proved for the last two:

sage: L2=[u.args[0].args[1]._sage_() for u in c_p.solve(theta, algorithm="sympy")]
sage: [c_p.subs(theta==u).trig_reduce().n() for u in L2]
[2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16,
 2.22044604925031e-16]

sage: [bool(c_p.subs(theta==u).trig_reduce()==0) for u in L2]
[False, False, True, True]

sage: var("k", domain="integer")
k
sage: LM=[u[1][2][1].sage(locals={"C":lambda x:k}) for u in mathematica.Solve(c_p
....: ==0, theta)]
sage: LM
[pi + 2*pi*k + arcsin(1/4*(2*pi + 5)/pi),
 2*pi*k - arcsin(1/4*(2*pi + 5)/pi),
 pi + 2*pi*k - arcsin(1/4*(2*pi - 5)/pi),
 2*pi*k + arcsin(1/4*(2*pi - 5)/pi)]

sage: [c_p.subs(theta==u).trig_reduce().expand() for u in LM]
[0, 0, 0, 0]