my ugly attempt

$\displaystyle cos(2x-\pi/2) =\sqrt2/2$

$\displaystyle cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2$

$\displaystyle (2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2$

/edit got another step

$\displaystyle (2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2$

$\displaystyle (2sinxcosx)=\sqrt2/2$