# Homework Help: Problem in finding a general solution

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1. Jul 13, 2018 at 12:58 PM

### navneet9431

1. The problem statement, all variables and given/known data

2. Relevant equations
General Formula for Tan(a)=Tan(b)

3. The attempt at a solution
See the question I have uploaded.

I have tried solving it this way,

Firstly I applied the Quadratic Formula to get,

Now we have two cases,

CASE-1

When

So General Formula here will be,

Now, CASE-2

when

So General Formula here will be

I do not know what should I do next to get the answer? Please tell me how to proceed Further.

The answer given in the key is the option (C).

I will be thankful for any help!

2. Jul 13, 2018 at 3:28 PM

### BvU

Hello,

So you need to combine $\pi (n+ {1\over 12})$ and $\pi (n - {5\over 12})$ . Leave the $\pi$ outside the brackets and try a few n. The pattern emerges !

3. Jul 13, 2018 at 3:57 PM

### Delta²

if the 6n looks scary in the option C, it is actually just $n/2$ cause $a=(6n+1)\frac{\pi}{12}=\frac{n}{2}\pi+\frac{\pi}{12}$. From this very last expression for a, what do you get if you put
1) n=even=2k
2) n=odd=2k+1

4. Jul 14, 2018 at 1:42 AM

### ehild

5. Jul 15, 2018 at 7:11 AM

### ehild

I think the problem is solved by @Delta2's hint, so I may come with a very simple solution of the equation tan2(α)+2√3 tan(α)=1,which can be rearranged to 1-tan2(α)=2√3 tan(α).
nπ/2 in the offered solutions suggests to solve the equation for 2α. The double-angle formula is $\tan(2α)=\frac{2\tan(α)}{1-\tan^2(α)}$, that is $\tan(2α)=\frac{2\tan(α)}{2\sqrt3 \tan(α)}=\frac{1}{\sqrt 3}$, that is, 2α=π/6+kπ and α=π/12+kπ/2.

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