How to solve $3\sec(x)-2\cot(x)>0$ (trigonometric inequality)? I am struggling to find the solution, here is what I already tried (it's for a pre-calculus class so no calculus):
$3\sec(x)-2\cot(x)=0 \rightarrow 3(\frac{1}{\cos(x)})=2\left(\frac{\cos(x)}{\sin(x)}\right) \rightarrow 3\sin(x)=2\cos(x)^2 \rightarrow 3\sqrt{1-\cos(x)^2}=2\cos(x)^2 \rightarrow 9(1-\cos(x)^2)=4\cos(x)^4 \rightarrow 4\cos(x)^4+9\cos(x)^2-9=0$
Then I found the roots ($\pm \sqrt{3}/2$) and the respective $rad$ values for $x$ ($\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6},\frac{7\pi}{6})$, but the problem is that when I graph it I find that the functions are different, although they share the roots (but cosine one has more roots in between!) and some patterns (maximum/minimum values and the roots of the cosine function seem to have some relation to when the original one is positive or negative).
Did I do the transformations right? I though they were supposed to be equal at least on the roots, why are they related in such a weird way?
But anyway, I think my biggest problem right now is that I have no intuition for associating the roots to the original function so that I know when its positive or negative.
Edit: The solution set is for all $\mathbb{R}$.
Edit 2 : Thank you all. So I did it using $\cos(x)^2=1-\sin(x)^2$ and now the roots are right($\frac{\pi}{6},\frac{5\pi}{6}$). My only problem now is associating this with the original function. I don't know how to aproach it. Looking at the graph, the tangent and the secant parts of the function seem to be independent for some reason.
 A: $ 3 \sec x - 2 \cot x \gt 0 $
Implies
$ \dfrac{3}{\cos x} \gt 2 \dfrac{\cos x }{\sin x} $
Multiply through by $\sin^2 x \cos^2 x $
$ 3 \cos x \sin^2 x \gt 2 \sin x \cos^3 x $
Hence,
$ \cos x \sin x ( 3 \sin x - 2 \cos^2 x ) \gt 0 $
But $\cos^2 x = 1 - \sin^2 x $, so
$ \cos x \sin x (3 \sin x + 2 \sin^2 x - 2 ) \gt 0 $
The quadratic in $\sin x$ factors into $( 2 \sin x - 1 )( \sin x + 2 ) $
Hence, we now have
$ \cos x \sin x (2 \sin x - 1) (\sin x + 2) \gt 0 $
The last term is always positive, so this reduces to
$\cos x \sin x (2 \sin x - 1) \gt 0 $
The zeros of the above function are $0, \dfrac{\pi}{6}, \dfrac{\pi}{2},\dfrac{5 \pi}{6}, \pi, \dfrac{3\pi}{2} $
Considering all three terms, the set where the inequality is satisfied is
$ S = (\dfrac{\pi}{6}, \dfrac{\pi}{2} ) \cup (\dfrac{5 \pi}{6}, \pi ) \cup (\dfrac{3\pi}{2}, 2 \pi) $
A: \begin{align*}
3\sec x - 2\cot x & > 0\\
\frac{3}{\cos x} - \frac{2\cos x}{\sin x} & > 0\\
\frac{3\sin x - 2\cos^2x}{\sin x\cos x} & > 0\\
\frac{3\sin x - 2(1 - \sin^2x)}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 3\sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 4\sin x - \sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin x(\sin x + 2) - 1(\sin x + 2)}{\sin x\cos x} & > 0\\
\frac{(2\sin x - 1)(\sin x + 2)}{\sin x\cos x} & > 0
\end{align*}
The term $\sin x + 2 > 0$ for every real value of $x$.  Thus, the sign of the expression depends on the terms $2\sin x - 1$ and $\sin x\cos x$.
In the interval $[0, 2\pi)$, the roots of the equation $2\sin x - 1 = 0$ are $\dfrac{\pi}{6}, \dfrac{5\pi}{6}$.  By examining the sine curve, we see that in the interval $[0, 2\pi)$, $2\sin x - 1 > 0 \implies \dfrac{\pi}{6} < x < \dfrac{5\pi}{6}$.
The term $\sin x\cos x > 0$ when $\sin x$ and $\cos x$ have the same sign, which occurs in the first and third quadrants.  The term $\sin x\cos x < 0$ when $\sin x$ and $\cos x$ have opposite signs, which occurs in the second and fourth quadrants.
We require that $2\sin x - 1 > 0$ and $\sin x\cos x > 0$ or that $2\sin x - 1 < 0$ and $\sin x\cos x < 0$.
In the interval $[0, 2\pi)$, $2\sin x - 1 > 0$ and $\sin x\cos x > 0$ if $\dfrac{\pi}{6} < x < \dfrac{\pi}{2}$.
In the interval $[0, 2\pi)$, $2\sin x - 1 < 0$ and $\sin x\cos x < 0$ if
$\dfrac{5\pi}{6} < x < \pi$ or $\dfrac{3\pi}{2} < x < 2\pi$.
Hence, the inequality is satisfied if
$$x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$$
or if $x$ is any angle coterminal with one of these angles.
