Trig equations solution Solving the following for x:
$$
\frac{3\cos(2x)+5\cos(x)-1}{\sqrt{-\cot(x)}}=0
$$
The solution says that the answer is $x=-\frac{\pi}{3}+2\pi k$ where $k$ is an integer.
I am not sure why there is the minus sign in front. Can someone please help me out?
 A: You're solving $\cos{x}=\frac{1}{2}$ and $\tan{x} = 0$.
So $x = \pm\frac{\pi}{3} + 2\pi k$ and $x = n\pi$
A: First $\sqrt{-\cot x} \neq 0$, then $-\cot x >0$, so $x\in A= ]\frac{\pi \;(2k+1)}{2},\pi(k+1)[\;\;\;\;\; \forall k\in \mathbb{Z}$. 
\begin{eqnarray}
3\cos 2x + 5\cos x -1 &=& 0\\
3(2\cos^{2}x-1)+5\cos x -1 &=&0\\
6\cos^{2}x+5\cos x -4&=&0\\
x_{1}=-\frac{\pi}{3}+2\pi k&,& x_{2}=\frac{\pi}{3}+2\pi k \text{$\;\; \forall k\in \mathbb{Z}$}
\end{eqnarray}
Now $x_{1}\in A$ but $x_{2}\notin A$, so, the solution of equation is $x_{1}=-\frac{\pi}{3}+2\pi k,\; \forall k\in \mathbb{Z}$
A: First off, there is a domain issue with the function on the left-hand side of the equation, since the denominator is undefined wherever $ \ \cot x \ \ge \ 0 \ . $  Since the tangent function is positive in the first and third quadrants, so is the cotangent function; also, $ \ \cot x \ = \ 0 \ $ wherever $ \ \cos x \ = \  0 $ and undefined where $ \ \sin x \ = \  0 $ .    Putting this all together tells us that there are only solutions to the equation for  $ \ (\frac{2n-1}{2}) \cdot \pi \ < \ x \ < \ n \cdot \pi \ , $ the second and fourth quadrants without their boundaries.
The numerator is zero for 
$$ \ 3 \ \cos(2x) \ + \ 5  \cos x \ - \ 1 \ = \ 3 \ (2 \cos^2 x \ - \ 1 ) + \ 5  \cos x \ - \ 1   $$
$$ = \ 6 \cos^2 x \ + \ 5  \cos x \ - \ 4 \ = \ (3 \ \cos x \ + \ 4 ) \ \cdot \ ( 2 \ \cos x \ - \ 1) \ = \ 0 .  $$
The first factor is never zero, but the second one is for $ \ \cos x \ = \ \frac{1}{2} \ \Rightarrow \ x \ = \ \pm \frac{\pi}{3}  +  2k\pi \ , $ as Oliver has said.  However, the $  + \frac{\pi}{3}  +  2k\pi \ $ "family" of solutions is barred since those are in the first quadrant.  Thus, the permissible family of solutions is  $  - \frac{\pi}{3}  +  2k\pi \ . $ 
