Set of $x$ values

Question

The set of all $x \in (0,\pi)$ satisfying $$1 +\log_2\sin x +\log_3\sin x\geq0$$ is$\ldots$

I tried to sketch the graph manually but could not complete it. I tried it in desmos and got the following graph. Is there any way to solve this question without looking at the computerised graph? Thanks.

Answer 1

Converting to natural logarithms (the only ones I know), you are concerned by the function $$f(x)=1+\frac{\log (\sin (x))}{\log (2)}+\frac{\log (\sin (x))}{\log (3)}=1+\frac{\log (6)}{\log (2) \log (3)} \log (\sin (x))$$ which has vertical asymptotes at $x=0$ and $x=\pi$.

Consider the derivative $$f'(x)=\frac{\log (6) }{\log (2) \log (3)}\cot (x)$$ It cancels when $x=\frac \pi 2$ and $$f\left(\frac \pi 2\right)=1$$ So, there are two $x$ intercepts on each side of $\frac \pi 2$.

The solutions corresponding to $$f(x)=0\implies \log( \sin(x))=-\frac{\log (2) \log (3)}{\log (6)}\implies\sin(x)=e^{-\frac{\log (2) \log (3)}{\log (6)}}$$ that is to say $$x_1=\sin ^{-1}\left(e^{-\frac{\log (2) \log (3)}{\log (6)}}\right)\approx 0.712555$$ and because of the symmetry $x_2=\pi-x_1\approx 2.42904$.

So, for any $x$ such that $x_1 < x < x_2$, $f(x)>0$.

Answer 2

write you inequality in the form $$1+\frac{\ln(\sin(x))}{\ln(2)}+\frac{\ln(\sin(x))}{\ln(3)}\geq 0$$ and you can solve for $$\sin(x)$$