Question 1116246
You could,
but it seems more direct to think of
{{{cos(2x)=0}}} --> {{{2x=pi/2+k*pi}}} for any {{{k}}} integer,
and from there solve for {{{x}}} as
{{{x=pi/4+k*pi/2}}}  for any {{{k}}} integer.
 
One of your alternatives is
{{{cos(2x)=0}}} --> {{{2cos^2(x)-1=0}}} --> {{{cos^2(x)=1/2}}} --> {{{cos(x)=" " +- sqrt(2)/2}}} --> {{{x=pi/4+k*pi/2}}}  for any {{{k}}} integer.