What is the set of solutions to $2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$ where $0<x<\pi$

Question

$$2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$$ where $0<x<\pi$

My process:

Since both $\sin x$ and $\cos x$ must be greater than $0$ and less than $1$, the initial set shrinks to $(0,\pi/2)$

Then after transformations I get

$2\log_{\cos x}\sin x+1-\log_{\sin x}\cos x\le 0$

. . .

$2t^2+t-1\le 0$

. . .

$\log_{\cos x}\sin x\in [-1,1/2] $

Can't be $[-1,0)$

But I'm not sure what the interval is further

I'd venture a guess as to say that $x\in (0,\frac{\sqrt5-1}{2}]$ but that doesn't fit with the solution on the exam, which doesn't explicitly give the solution but says it is in the form of $(a,b)$

Any help appreciated.

Answer 1

We need $\displaystyle 0<x<\frac{\pi}{2}$. Note that $\ln\cos x<0$ and $\ln \sin x<0$.

\begin{align} \frac{2\ln \sin x}{\ln \cos x}&\le\frac{\ln \cos x-\ln\sin x}{\ln\sin x}\\ 2(\ln\sin x)^2&\le(\ln\cos x)^2-\ln\sin x\ln \cos x\\ (\ln\cos x)^2-\ln\sin x\ln \cos x-2(\ln\sin x)^2&\ge0\\ (\ln \cos x+\ln \sin x)(\ln\cos x-2\ln \sin x)&\ge0\\ \ln\cos x-2\ln\sin x&\le0\\ \ln\cos x&\le\ln\sin^2x\\ \cos x&\le1-\cos^2x\\ \left(\cos x+\frac{1}{2}\right)^2&\le\frac{5}{4}\\ 0<\cos x&\le\frac{-1+\sqrt{5}}{2}\\ \cos^{-1}\left(\frac{-1+\sqrt{5}}{2}\right)\le x&<\frac{\pi}{2} \end{align}

Answer 2

We can rewrite this equation using the change of base formula:

$$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$$ for any value of $c$.

Here, we can say that $c=e$, i.e. we are using the natural logarithm:

\begin{align}2\log_{\cos x}\sin x&\leq \log_{\sin x}\cot x\\\\ \frac{2\ln(\sin x))}{\ln(\cos(x))}&\leq \frac{\ln(\cot(x))}{\ln(\sin(x))}\end{align}

Can you now solve this?

Answer 3

The first restriction to $0<x<\pi/2$ is correct.

Note now that $$ \log_ab=\frac{1}{\log_ba} $$ so if you set $t=\log_{\sin x}\cos x$ you get $$ \frac{2}{t}+1-t\le0 $$ that is $$ \frac{t^2-t-2}{t}\ge0 $$ The solution set of this is $$ [-1,0)\cup[2,\infty] $$ Can you finish?