Problem in finding a general solution

[navneet9431]

Homework Statement

Homework Equations

General Formula for Tan(a)=Tan(b)

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!

 

[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 !

 

[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

 

[ehild]

Homework Statement

View attachment 228016

Remember the double-angle formulas.
http://mathworld.wolfram.com/Double-AngleFormulas.html

What do you get for tan(2α) using the original equation tan²(α)+2√3 tan(α)=1 ? What is the general solution for 2α?

 

[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.