How do you solve #2tanA=-sec^2A#?

It's possible to deal with a trigonometric equation using substitution if all terms use the same trigonometric function .

Starting with #2tanA = -sec^2 A#, we can use the identity:
#tan^2A + 1 = sec^2A#

to convert the #sec# term into a #tan# term.

#2tanA = -sec^2 A#
#2tanA = -(tan^2A+1)#

Then, we can rearrange the terms:

#tan^2A +2tanA+1=0#

This might look similar to a quadratic equation, and it can be solved in the same manner. The key step is to let a variable represent the trigonometric function, then solve for the values of that variable:

#tan^2A +2tanA+1=0#
Let #k = tanA#.
#k^2 + 2k + 1 = 0#
#(k+1)^2=0#

The solution to this equation is #k=-1#.

Then, we use our earlier substitution to solve for our original variable, #A#.

#k=-1#, #k = tanA#

#tan A = -1#

The tangent function has a period of #pi#, so let's first consider the solutions from #[0,pi]#. There is only one such solution:

#tan((3pi)/4)=-1#

Since the tangent function has a period of #pi#, the solution repeats every #pi# units.
Thus, the general solution is:
#{A|A = (3pi)/4 + npi, n in ZZ}#