How do you solve sin 2x/sin x = 1+cos 2x/cos 2x?

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Answer 1 · 2018-05-03T06:11:10

#x=(2n+1)pi/2# or #x=2npi+-1.9446# in radians

#(sin2x)/sinx=(1+cos 2x)/(cos2x)#

#=>(2sinxcosx)/sinx=(2cos^2x)/(2cos^2x-1)#

or #cosx(2cos^2x-1)-2cos^2x=0#

or #cosx(2cos^2x-1-2cosx)=0#

i.e. either #cosx=0# i.e. #x=(2n+1)pi/2#

or #2cos^2x-2cosx-1=0#

i.e. #cosx=(2+-sqrt(4+8))/4=(1+-sqrt3)/2#

as we cannot have #cosx# as #(1+sqrt3)/2#, it being more than #1#, only possibility is #cosx=(1-sqrt3)/2#

and #x=2npi+-1.9446# in radians