How do you prove that #cos 2x(1 + tan 2x) = 1#?

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Answer 1 · 2015-02-02T11:22:54

The answer2 are: #x=kpi# and #pi/4+kpi#.

The equation can be written:

#cos2x+cos2xtan2x=1rArrcos2x+cos2x(sin2x)/(cos2x)=1rArrcos2x+sin2x=1rArrsin2x+cos2x=1#.

Now it is possible multiply both members for #sqrt2/2#:

#sqrt2/2sin2x+sqrt2/2cos2x=sqrt2/2rArr#

#sin2xcos(pi/4)+cos2xsin(pi/4)=sqrt2/2#.

Using the addition formula:

#sin(2x+pi/4)=sqrt2/2#.

The sinus is #sqrt2/2# if its argument is #pi/4+2kpi# or #3/4pi+2kpi#.

So:

#2x+pi/4=pi/4+2kpirArr2x=2kpirArrx=kpi#,

and

#2x+pi/4=3/4pi+2kpirArr2x=pi/2+2kpirArrx=pi/4+kpi#.