Find the solutions of `2 sin^2 x = sqrt(3) sin 2x` that are at least `0` and less than `2pi`. equation x=0 \/ x=pi/3 \/ x=pi \/ x=4pi/3 ends /* `x` must be at least `0` and less than `2 pi` */ f_nodes 25 0 <= x < 2pi /\ 2 sin^2 x = sqrt(3) sin 2x /**/
<=> end_of_answer next_URL http://math.tut.fi/mathcheck/equation5.html
or
This file was generated 2017-06-09 08:36:02 UTC.