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How do you solve for x in: #tan^2 x - 3 tan x - 4 = 0#?

1 Answer

Apr 15, 2015

Let #k=tan(x)#
then
#tan^2(x) -3tan(x)-4 = 0#
becomes
#k^2-3k-4=0#

#(k-4)(k+1) = 0#

So we are looking for
#k=4# or #k=-1#
that is, for
#tan(x) = 4# or #tan(x)=-1#

#x = arctan(4)#
or
#x= arctan(-1) = -pi/4 +n(pi)# for all #n epsilon ZZ#

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