Question #7d61a

1 Answer
May 13, 2017

x=pi/4+(npi)/2 where ninZZ

Explanation:

Question: sec^2x+tan^2x=3tan^2x

We can move the tan^2x to the right-hand side:
sec^2x=2tan^2x

Note that secx=1/cosx and tanx=sinx/cosx
So we have:
1/cos^2x=(2sin^2x)/cos^2x
1/cos^2x-(2sin^2x)/cos^2x=0
(1-2sin^2x)/cos^2x=0

Note the double angle identity for cos(2x)=cos^2x-sin^2x=1-2sin^2x
So we have:
(cos^2x-sin^2x)/cos^2x=0

cos^2x/cos^2x-sin^2x/cos^2x=0

1-sin^2x/cos^2x=0

1-tan^2x=0

tan^2x=1

tanx=+-sqrt1

tanx=+-1

Therefore,
x=pi/4+(npi)/2 where ninZZ