How do you find the general solution of #sqrt2 sin2theta+1=0#?

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Answer 1 · 2018-06-24T06:16:03

#theta=-pi/8+npi# or #theta=-(3pi)/8+npi# where n is an integer

#sqrt2sin2theta+1=0#

#sqrt2sin2theta=-1#

#sin2theta=-1/sqrt2#

Recall that #sin2theta# is negative in both the 3rd and 4th quadrant

#2theta=-pi/4,-(3pi)/4,...#

#2theta=-pi/4+2npi# or #2theta=-(3pi)/4+2npi# where n is an integer

#theta=-pi/8+npi# or #theta=-(3pi)/8+npi#

Answer 2 · 2018-06-24T06:19:29

The solutions are #S={5/8pi+kpi, 7/8pi+kpi,}#, #k in ZZ#

The equation is

#sqrt2sin2theta+1=0#

#=>#, #sqrt2sin2theta=-1#

#=>#, #sin2theta=-1/sqrt2#

Therefore,

#{(2theta=5/4pi+2kpi),(2theta=7/4pi+2kpi),(2theta=13/4pi+2kpi),(2theta=15/4pi+2kpi):}#

#<=>#, #{(theta=5/8pi+kpi),(theta=7/8pi+kpi),(theta=13/8pi+kpi),(theta=15/8pi+kpi):}#, #AA k in ZZ#