How do you solve #tanx=tan^2x# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-09-23T23:35:24

#x = 0, (pi) / (4), (3 pi) / (4), pi, (5 pi) / (4), (7 pi) / (4), 2 pi#

We have: #tan(x) = tan^(2)(x)#; #0 leq x leq 2 pi#

First, let's subtract #tan(x)# from both sides of the equation:

#=> tan^(2)(x) - tan(x) = 0#

Then, let's take #tan(x)# in common:

#=> tan(x) (tan(x) - 1) = 0#

Now that we have a product that is equal to zero, either one of the multiples must be equal to zero:

#=> tan(x) = 0#

#=> x = 0, (pi - 0), (pi + 0), (2 pi - 0)#

#=> x = 0, pi, 2 pi#

or

#=> tan(x) - 1 = 0#

#=> tan(x) = 1#

#=> x = (pi) / (4), (pi - (pi) / (4)), (pi + (pi) / (4)), (2 pi - (pi) / (4))#

#=> x = (pi) / (4), (3 pi) / (4), (5 pi) / (4), (7 pi) / (4)#

Therefore, the solutions to the equation are #x = 0, (pi) / (4), (3 pi) / (4), pi, (5 pi) / (4), (7 pi) / (4), 2 pi#.