# Math Help - General solutions.

1. ## General solutions.

The question is to find the general solution for the first line of the following picture:

My method is included in that to but it differs from the book answer. One of the biggest problems is that I have two answers and they only have one!

I figured that the second solution must be included in the first, but for my solutions this doesn't happen.

Can anyone point out what i'm doing wrong?

2. Your answer is as correct as can be.Try taking LCM and you will get a factor as $4n+1$ and in the other you will get $4n+3$.By giving different integral values to $n$ you will find that they comprise of the same values as you will obtain when you will put them in $2n+1$

Another method can be
$\tan 5\theta=\cot 4\theta = \tan (\frac {\pi}{2}-4\theta)$

$5\theta=n\pi +\frac {\pi}{2}-4\theta$

$9\theta=n\pi +\frac{\pi}{2}$

$\theta=\frac{n\pi}{9} +\frac{\pi}{18}$

$\theta=(2n+1)\frac{\pi}{18}$

3. I sort-of-get what you've done, except the first line. Where does that come from?

I'll also try and get their answer from mine.

4. Originally Posted by Showcase_22
I sort-of-get what you've done, except the first line. Where does that come from?

I'll also try and get their answer from mine.
$\cot A = \tan \left( \frac{\pi}{2} - A\right)$.

The proof is simple:

RHS $= \tan \left( \frac{\pi}{2} - A\right) = \frac{\sin \left( \frac{\pi}{2} - A\right)}{\cos \left( \frac{\pi}{2} - A\right)} = \frac{\cos A}{\sin A} = \cot A =$ LHS.