• Apr 19th 2012, 06:27 PM
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.
• Apr 19th 2012, 06:53 PM
Prove It
Quote:

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.

The period of the tangent function is \displaystyle \displaystyle \begin{align*} \pi \end{align*}, so the solution to the first equation becomes

\displaystyle \displaystyle \begin{align*} \tan{3x} + 1 &= 0 \\ \tan{3x} &= -1 \\ 3x &= \arctan{(-1)} + \pi k \textrm{ where }k \in \mathbf{Z} \end{align*}

What happens when you divide this all by \displaystyle \displaystyle \begin{align*} 3 \end{align*} to find \displaystyle \displaystyle \begin{align*} x \end{align*}?

Now follow a similar process for \displaystyle \displaystyle \begin{align*} \tan{3x} - 1 = 0 \end{align*}.
• Apr 19th 2012, 10:15 PM
biffboy