Question #6f1a1

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Answer 1 · 2017-08-10T14:48:51

The solutions are #a={pi/4+2kpi, 3/4pi+2kpi,5/4pi+2kpi, 7/4pi+2kpi}# and #b={pi/6+2kpi, 5/6pi+2kpi, 7/6pi+2kpi, 11/6pi+2kpi}#, #AA k in ZZ#

#tana/tanb=sina/sinb*cosb/cosa=sqrt3#

#sina/sinb=sqrt2#, #=>#, #sina=sqrt2sinb#

Therefore,

#cosb/cosa=sqrt(3/2)#, #=>#, #cosa=sqrt(2/3)cosb#

#sin^2a+cos^2a=2sin^2b+2/3cos^2b=1#

#2sin^2b+2/3(1-sin^2b)=1#

#2sin^2b+2/3-2/3sin^2b=1#

#4/3sin^2b=1/3#

#sin^2b=1/4#, #=>#, #sinb=+-1/2#

#b={pi/6+2kpi, 5/6pi+2kpi, 7/6pi+2kpi, 11/6pi+2kpi}#, #AA k in ZZ#

#sina=sqrt2*(+-1/2)=+-sqrt2/2#

#a={pi/4+2kpi, 3/4pi+2kpi,5/4pi+2kpi, 7/4pi+2kpi}#, #AA k in ZZ#