Simplify trigonometric expression

[courteous]

Simplify: $\displaystyle \frac{2\sin{x}+1}{\sqrt3 - 2\cos{x}}$

What I've come to thus far (by multiplying with $\displaystyle \frac{\sqrt3 + 2\cos{x}}{\sqrt3 + 2\cos{x}}$ and $\displaystyle \frac{2\sin{x}-1}{2\sin{x}-1}$) is $\displaystyle \frac{\sqrt3 + 2\cos{x}}{2\sin{x}-1}$ :x

Oh, not to forget, the solution deals with $\displaystyle \cot$ (cotangens).

 

[Deleted member 4993]

Simplify: $\displaystyle \frac{2\sin{x}+1}{\sqrt3 - 2\cos{x}}$

What I've come to thus far (by multiplying with $\displaystyle \frac{\sqrt3 + 2\cos{x}}{\sqrt3 + 2\cos{x}}$ and $\displaystyle \frac{2\sin{x}-1}{2\sin{x}-1}$) is $\displaystyle \frac{\sqrt3 + 2\cos{x}}{2\sin{x}-1}$ :x

Oh, not to forget, the solution deals with $\displaystyle \cot$ (cotangens).

Please share with us the "exact" form you are given as answer.

Without knowing the form that is wanted - I would do it the following way

$\displaystyle \frac{2sin(x) + 1}{\sqrt{3}-2cos(x)}$

$\displaystyle = \, \frac{sin(x) + \frac{1}{2}}{\frac{\sqrt{3}}{2}-cos(x)}$

$\displaystyle = \, \frac{sin(x) + \, sin(\frac{\pi}{6})}{cos(\frac{\pi}{6}) \, - \, cos(x)}$

then use the fact that:

$\displaystyle sin(A) \, + \, sin(B) \, = \, 2\cdot sin(\frac{A+B}{2})\cdot cos(\frac{A-B}{2})$

and similar form for cosines.

That should take you to the "allegedly simlified" cotangent form.

 

[courteous]

With sum as product, I managed to take it to given solution $\displaystyle \cot\frac{x-\frac{\pi}{6}}{2}$.