Question #6d434

1 Answer
Dec 15, 2017

#x=pi/8,(3pi)/8,(9pi)/8,(11pi)/8#

Explanation:

#sqrt2sin2x=1#

#rArrsin2x=1/sqrt2#

#"note "0<2x<4pi#

#rArr2x=pi/4,(pi-pi/4),(pi/4+2pi),((3pi)/4+2pi)#

#color(white)(rArr2x)=pi/4,(3pi)/4,(9pi)/4,(11pi)/4#

#rArrx=pi/8,(3pi)/8,(9pi)/8,(11pi)/8to(0,2pi)#