How can this be verified?

Question

#2sin2x - sqrt2 = 0#

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Answer 1 · 2018-03-24T22:28:18

Will solve it...
#x= pi/8+pin#
#x= (3pi)/8+pin#

n is an element of all integers

#2sin2x-sqrt2=0#

#2sin2x=sqrt2#

#sin2x=sqrt2/2#

#2x= pi/4+2pin#
#2x= (3pi)/4+2pin#

#x= pi/8+pin#
#x= (3pi)/8+pin#

To confirm, the graph:
graph{2sin(2x)-sqrt2 [-6.755, 13.195, -6.09, 3.89]}