How do I solve for x in 4(sin(x-pi))(cos(x-pi)) = sqrt2 ?

1 Answer
Mar 21, 2018

#x in {(k+9/8)pi, (k+11/8)pi|kinZZ} #

Explanation:

#4sin(x-pi)cos(x-pi)=sqrt2#

#rArr2sin(x-pi)cos(x-pi)=sqrt2/2#

The LHS is now in a standard form, so we use the following identity

#2sinxcosx=sin2x#

#rArr2sin(x-pi)cos(x-pi)=sin(2(x-pi))=sqrt2/2#

#rArr2(x-pi)= 1/4pi+2kpi# for # k in ZZ#

Or #2(x-pi)=3/4pi +2kpi#

#rArr x-pi=1/8pi+kpi rArr x=9/8pi+kpi=(k+9/8)pi#

Or #x-pi=3/8pi+kpi rArr x=11/8pi+kpi=(k+11/8)pi#