trig equation help please

Hello, I am trying to solve this:

tan^2(3x)=1 (for values between 0 and 2pi)

I have broken it down to a 'difference of squares' format:

(tan3x+1) (tan3x-1)

I then find the inverse tan of 1 and -1 which gives me 45 degrees and -45 degrees.
I then divide them by three to give me plus/minus 15.

I understand that is the first part of the answer, however my text book says the complete answer is +-pi/12 + pi/3 K.
I understand the pi/3 K part is for repetition, but that is only repeating it every 60 degrees. Looking at graph of tan I see the repetitions would be every 90 degrees ?

Thanks for any help.

Quote:

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Originally Posted by fran1942

Hello, I am trying to solve this:

tan^2(3x)=1 (for values between 0 and 2pi)

I have broken it down to a 'difference of squares' format:

(tan3x+1) (tan3x-1)

I then find the inverse tan of 1 and -1 which gives me 45 degrees and -45 degrees.
I then divide them by three to give me plus/minus 15.

I understand that is the first part of the answer, however my text book says the complete answer is +-pi/12 + pi/3 K.
I understand the pi/3 K part is for repetition, but that is only repeating it every 60 degrees. Looking at graph of tan I see the repetitions would be every 90 degrees ?

Thanks for any help.

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The period of the tangent function is , so the solution to the first equation becomes



What happens when you divide this all by to find ?

Now follow a similar process for .

You could have gone straight from tan^2 3x=1 to saying tan3x=+1 or -1