How do you solve #sin x cos x = [sqrt 2]/4#?

I would multiply and divide by #2# on the left:
#color(red)(2/2)sin(x)cos(x)=sqrt(2)/4#
rearranging and using a trig identity:
#2sin(x)cos(x)=2sqrt(2)/4#
#color(red)(sin(2x))=cancel(2)sqrt(2)/cancel(4)^2#
#2x=pi/4 or 3/4pi#
and
#x=pi/8 or3/8pi#
for #x# in the interval: #0<=##x##<=2pi#

Suppose #h = sin(x)#
then
#color(white)("XXXX")##cos(x) = sqrt(1-h^2)#

#sin(x)*cos(x)= sqrt(2)/4# is equivalent to
#color(white)("XXXX")##h(sqrt(1-h^2))= sqrt(2)/4#

#color(white)("XXXX")##sqrt(1-h^2) = sqrt(2)/(4h)#

#color(white)("XXXX")##1-h^2 = 2/(16h^2)#

#color(white)("XXXX")##16h^4-16h^2+2=0#

Let #k=h^2#

#color(white)("XXXX")##8k^2-8k+1=0#

#color(white)("XXXX")##k= (8+-sqrt(64-32))/16#

#color(white)("XXXX")##color(white)("XXXX")##=(2+sqrt(2))/4 or (2-sqrt(2))/4#

#color(white)("XXXX")##h = sqrt((2+sqrt(2))/4) or sqrt((2-sqrt(2)/4)#
(using my handy calculator)
#color(white)("XXXX")##h = 0.92388 or 0.382683#

#x = arcsin(h)#
(and again with my calculator)
#color(white)("XXXX")##x=(3pi)/8 or pi/8#

Given that #x# came out as such pretty values, I expect there is a better (prettier) way to do this.