How do you find the general solution of #\sec ^ { 2} 2x = 1- \tan 2x#?

1 Answer
maganbhai P.
Mar 16, 2018

#x=(kpi)/2,kinZorx=(kpi)/2-pi/8,kinZ#

Explanation:

We know that,
#tantheta=tanalpha=>theta=kpi+alpha,kinZ#
Here,
#color(red)(sec^2(2x))=1-tan(2x)#
#=>color(red)(1+tan^2(2x))=1-tan(2x)#
#=>tan^2(2x)=-tan(2x)#
#=>tan^2(2x)+tan(2x)=0#
#=>tan2x(tan2x+1)=0#
#=>tan2x=0ortan2x=-1#
#=>tan2x=tan0ortan2x=tan(-pi/4)#
#=>2x=kpi,kinZor2x=kpi-pi/4,kinZ#
#:.x=(kpi)/2,kinZorx=(kpi)/2-pi/8,kinZ#

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