The statement of the problem : Solve in $\mathbb R$ the following equation : $$\frac{\sin3x + 2}{\sin x+2} = 2023^{\sin x-\sin3x}$$
My approach : To simplify, we will use the fact that $\sin3x = 3*\sin x -4*\sin^3 x$ . So, if we make the notation $t=\sin x$ , where $t \in [ -1 , 1 ]$ we get the following : $$\frac{-4t^3+3t+2}{t+2} = 2023^{4t^3 – 2t}$$
Now I tried for some time to find some solutions and below you can see a graph, but it seems to me that it is getting super complicated, I managed to find some solutions, but it is difficult for me to prove that they are the only ones.
If possible, I would prefer a solution that provides the use of inequalities or convexity/concavity or inverse trigonometric functions rather than the use of limits or derivatives because I have to present it to a group of students who have not yet learned about these. The problem is from an old textbook and it is before the lesson on limits and derivatives, so there is probably another solution that don't implies them which I'm looking forward to.
All proofs will be helpful. Thanks a lot!
Best Answer
Here is my proof: We have that $$\frac{\sin(3x)+2}{\sin(x)+2}=2023^{\sin(x)-\sin(3x)}$$ So $$2023^{\sin(3x)}(\sin(3x)+2)=2023^{\sin(x)}(\sin(x)+2)$$ we consider three cases:
$\bullet$If: $\sin(3x)>\sin(x)$
We can deduce that $2023^{\sin(3x)}>2023^{\sin(3x)}>0$ and $\sin(3x)+2>\sin(x)+2>0$
Then $$2023^{\sin(3x)}(\sin(3x)+2)>2023^{\sin(x)}(\sin(x)+2)$$ $\bullet$If: $\sin(3x)<\sin(x)$
We have the same absurd (by symmetry)
$\bullet$If: $\sin(3x)=\sin(x)$
The only solutions are verifying $3x=x+2k\pi$ or $3x=\pi-x+2k\pi$ with $k\in\mathbb{Z}$
And finally $x=k\pi$ or $x=\frac{\pi}{4}+k\frac{\pi}{2}$ with $k\in\mathbb{Z}$
In conclusion:
$$\mathbb{S}=\{k\pi,k\in\mathbb{Z}\}\cup\{\frac{\pi}{4}+k\frac{\pi}{2},k\in\mathbb{Z}\}$$