How do you find all solutions of the equation in the interval #[0,2pi)# given #2sin2x-sqrt2=0#?

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Answer 1 · 2017-11-18T16:20:22

#pi/8 and (9pi)/8#

#2sin(2x)-sqrt(2)=0#

#sin2x=sqrt(2)/2#

#2x=sin^-1(sin2x)=sin^-1(sqrt(2)/2)=pi/4#

#2x=pi/4=>x=pi/8 #

#-pi/8=pi+pi/8=(9pi)/8#

For: #color(white)(888)0 < x <=pi#

#pi/8 and (9pi)/8#