Solve (sinalpha/sin2x)+ (cosalpha/cos2x)=2?

#(sinalpha)/(sin2x)+ (cosalpha)/(cos2x)=2#

1 Answer
Mar 10, 2018

#=>x=(2k+1)pi/6-alpha/6,kinZorx=kpi+alpha/2,kinZ#

Explanation:

#(sinalpha)/(sin2x)+ (cosalpha)/(cos2x)=2##=>cos2xsinalpha+sin2xcosalpha=2sin2xcos2x#
#=>sin(2x+alpha)=sin4x#
#=>sin4x-sin(2x+alpha)=0#
#=>2cos((4x+2x+alpha)/2)sin((4x-2x-alpha)/2)=0#
#=>cos(3x+alpha/2)=0orsin(x-alpha/2)=0#
#=>3x+alpha/2=(2k+1)pi/2,kinZorx-alpha/2=kpi,kinZ#
#=>3x=(2k+1)pi/2-alpha/2,kinZorx=kpi+alpha/2,kinZ#
#=>x=(2k+1)pi/6-alpha/6,kinZorx=kpi+alpha/2,kinZ#