How do you solve 3sec^(2)x-2tan^(2)x-4=0?

#3sec^2x-2tan^2x-4=0#

1 Answer
Feb 20, 2018

By using some trigonometric identities.
If you wish to see only the solutions, scroll down until the end.

Explanation:

Let's factor out the #-2# :

#3sec^2 x - 2(tan^2 x + 2) = 0#.

We can rewrite #tan^2 x + 2# as #tan^2 x + 1 + 1#.
Now, #tan^2 x + 1# is equal to #sec^2 x#, as proved below :

#tan^2 x + 1 = sin^2 x/cos^2 x + 1#.

Instead of #1#, let's use #cos^2 x/cos^2 x#.

#tan^2 x + 1 = (sin^2 x + cos^2x)/cos^2 x#

#tan^2 x + 1 = 1/cos^2 x = sec^2 x#.

Back to the original expression;

#3sec^2 x - 2(sec^2 x + 1) = 0#

#3sec^2 x - 2 sec^2 x - 2 = 0#

#sec^2 x = 2#

#1/cos x = sqrt(2)#

#cos x = 1/sqrt(2)#.

The first value of #x# that comes to mind is #pi/4#, however, that relation is true for all numbers of the form :

#2npi+pi/4#
and
#2npi-pi/4#

where #n# is an integer.

In conclusion (or tl;dr) :

The set of solutions is

#S# = #{2npi+pi/4 | n -> "Z"} uu {2npi - pi/4 | n -> "Z"}#.