Dk $x\neq 0;\pi $$pt<=>\frac{2cos2x.sinx}{\sqrt{2}. \left| {sinx} \right|}=\sqrt{2}sin(2x+\frac{\pi }{4})$
$<=>2cos2x.sinx=2\left| {sinx} \right|.sin(2x+\frac{\pi }{4})$
Den day chi xet khoang pha tri tuyet doi thoi. viec con lai qua don gian vs ban oy.