Since, by definition, #sec(x)=1/cos(x)#, we can re-write the given equation as
#sin(x)/cos(x) = 2 sin(x)#
First of all, we should define the domain where this equation is defined. Since #cos(x)# is in the denominator, our domain should exclude from the given interval #[0,2pi]# points where #cos(x)=0#, that is we should restrict our solutions to these conditions:
#x!=pi/2# and #x!=(3pi)/2#
Next we notice that both parts of the equation, left and right, contain a factor #sin(x)# that can be divided upon to simplify the equation. But, by doing so, we have to make sure that we don't lose any solutions, in particular, those values of #x# where #sin(x)=0#. Obviously, they are the solutions of the original equation since both parts of it would be equal to zero, but they might not be among the solutions of the new equation after we reduce left and right parts by #sin(x)#.
So, from #sin(x)=0# we derive three solutions:
#x=0#, #x=pi# and #x=2pi#, all within a given interval #[0,2pi]# and none excluded from it by the logic above related to the domain of this equation.
Finally, after reduction by #sin(x)#, we obtain an equation
#1/cos(x)=2# or #cos(x)=1/2#,
which results in two more solutions:
#x=pi/3# and #x=2pi-pi/3=(5pi)/3#
ALWAYS DO THE CHECKING OF ALL SOLUTIONS OBTAINED
