How do you solve #sqrt2sin2x=1# between the interval #0<=x<=2pi#?

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Answer 1 · 2016-07-19T02:39:45

We have that

#sqrt2 sin2x=1=>sin2x=1/sqrt2=>sin2x=sqrt2/2=> sin2x=sin(pi/4)#

Hence from the last relation we get

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

or

#2x=2kpi+pi-pi/4=>2x=2kpi+3pi/4=>x=kpi+3pi/8#

Because #0<=x<=2pi# acceptable solutions are

#x_1=pi/8,x_2=9pi/8,x_3=3pi/8,x_4=11pi/8#