[Math] Logarithm Trigonometric Equation

Question

I am interested in solving the following question:
\begin{equation*}
\log _{\sin(x)\cos(x)}(\sin(x))\times \log_{\sin(x)\cos(x)}(\cos(x))=\frac{1}{4}
\end{equation*}
One approach that I tried was as follows; Let
\begin{equation*}
\log_{\sin(x)\cos(x)}(\sin(x))=p,~\log_{\sin(x)\cos(x)}(\cos(x))=q.
\end{equation*}
This implies that $pq=1/4$. Using the fact that $\log_a b=c$ can be written as $a^c=b$, we have
\begin{equation*}
\left(\sin(x)\cos(x)\right)^p=\sin(x),~\left(\sin(x)\cos(x)\right)^q=\cos(x).
\end{equation*}
Using the law $\sin^2(x)+\cos^2(x)=1$, we can write
\begin{equation*}
1=\sin^2(x)+\cos^2(x)=(\sin(x)\cos(x))^{2p}+(\sin(x)\cos(x))^{\frac{1}{2p}}=\frac{1}{4}
\end{equation*}
however I am not sure what to do from here (or if this is even the correct approach). I also tried substituting the equation in Wolfram alpha as well as but both returned nothing of use to me. I will be grateful if anyone can help advance this problem. Thanks in advance.

Answer 1

Rewrite using the natural logarithm,

$$\frac{\ln\sin x}{\ln(\sin x\cos x)}\frac{\ln\cos x}{\ln(\sin x\cos x)}=\frac14$$

or

$$(\ln\sin x+\ln\cos x)^2-4\ln\sin x\ln\cos x=0$$

or

$$(\ln\sin x-\ln\cos x)^2=0.$$

The rest is easy.