Vtyjkyuk

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

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

You basically finished and found the solution. Indeed:
$$cos x=frac12 Rightarrow x=pmfracpi3+2pi m,min mathbb Z;\
-6<x<0 Rightarrow -6<pmfracpi3+2pi m<0\
sin 2x>0 Rightarrow 2sin xcos x>0 Rightarrow sin x>0 Rightarrow \
0+2pi n<x<pi+2pi n,nin mathbb Z.$$
Hence:
$$begincases-6<pmfracpi3+2pi m<0\2pi n<x<pi+2pi nendcases Rightarrow begincases-6<-fracpi3;fracpi3-2pi<0\ -2pi<fracpi3-2pi<pi-2piendcases Rightarrow \
x=fracpi3-2pi=-frac5pi3.$$

First of all $dfrac-6x-x^210$ should be non-1 positive and $sin 2x>0$ and $sin x+sin 3x>0$ therefore $$x^2+6x<0\x^2+6xne -10$$but the latter inequality holds automatically since $Delta<0$. Also the answers of $sin 2x=sin x+sin 3x$ are$$sin x=0\textand/or\2cos x=4-4sin^2x=4cos^2x$$therefore $$x=kpi\x=kpi+dfracpi2\x=2kpipmdfracpi3$$the only valid answer is $$x=-dfrac5pi3$$