math
posted by mike .
how would you solve for x:
(3)^.5sin(2x)2cos(2x)+2sin^2(x)=1
using trig identites. please help

took me awhile, but here it is ...
√3sin(2x)  2cos(2x) = 1  2sin^2 x
√3sin(2x)  2cos(2x) = cos(2x)
√3sin(2x) = 3cos(2x)
sin(2x)/cos(2x) = 3/√3
tan(2x) = 3/√3
2x = 60º or 240º
then x = 30º or 120º
or
x = pi/6 or x = 2pi/3
I checked both answers, they work