Solve $\log_{(-x^2-6x)/10}(\sin 3x+\sin x)= \log_{(-x^2-6x)/10}(\sin 2x).$

Question

Question : Solve $$\log_{(-x^2-6x)/10}(\sin 3x+\sin x)= \log_{(-x^2-6x)/10}(\sin 2x).$$

My try :

I am unable to proceed from here

Answer 1

You basically finished and found the solution. Indeed: $$\cos x=\frac12 \Rightarrow x=\pm\frac{\pi}{3}+2\pi m,m\in \mathbb Z;\\ -6<x<0 \Rightarrow -6<\pm\frac{\pi}{3}+2\pi m<0\\ \sin 2x>0 \Rightarrow 2\sin x\cos x>0 \Rightarrow \sin x>0 \Rightarrow \\ 0+2\pi n<x<\pi+2\pi n,n\in \mathbb Z.$$ Hence: $$\begin{cases}-6<\pm\frac{\pi}{3}+2\pi m<0\\2\pi n<x<\pi+2\pi n\end{cases} \Rightarrow \begin{cases}-6<-\frac{\pi}{3};\frac{\pi}{3}-2\pi<0\\ -2\pi<\frac{\pi}{3}-2\pi<\pi-2\pi\end{cases} \Rightarrow \\ x=\frac{\pi}{3}-2\pi=-\frac{5\pi}{3}.$$

Answer 2

The base of the logarithms is the same, so you get $$ \begin{cases} \sin3x+\sin x=\sin2x \\[4px] \sin2x>0 \\[4px] (-x^2-6x)/10>0 \\[4px] (-x^2-6x)/10\ne 1 \end{cases} $$

Answer 3

First of all $\dfrac{-6x-x^2}{10}$ should be non-1 positive and $\sin 2x>0$ and $\sin x+\sin 3x>0$ therefore $$x^2+6x<0\\x^2+6x\ne -10$$but the latter inequality holds automatically since $\Delta<0$. Also the answers of $\sin 2x=\sin x+\sin 3x$ are$$\sin x=0\\\text{and/or}\\2\cos x=4-4\sin^2x=4\cos^2x$$therefore $$x=k\pi\\x=k\pi+\dfrac{\pi}{2}\\x=2k\pi\pm\dfrac{\pi}{3}$$the only valid answer is $$x=-\dfrac{5\pi}{3}$$