For $x\in (0,\pi)$, how many solutions does the equation $\sin x+2\sin2x-\sin3x=3$ have?
My attempt is as follows:
One can see that if $\sin3x$ is positive, then it is not possible to have any solution, so $\sin3x$ should be negative.
$$\sin3x<=0$$ If $x\in (0,\pi)$, then $3x \in(0,3\pi)$
So $\sin3x\in[-1,0]$
We can also see that $\sin x$ and $\sin 2x$ should be positive in order to have the solution.
$$\sin x>0 \text { and } \sin2x>0$$ As $x \in (0,\pi)$, so $\sin x\in(0,1]$
But $\sin2x>0$, so it means $x$ has to be acute (less than $\dfrac{\pi}{2}$)
So $x$ will be in first quadrant, $2x$ will be in second quadrant, $3x$ will be in second quadrant.
We can also see that if $\sin x$ is high then $\sin2x$ will be low because graph of $\sin x$ is increasing in first quadrant and decreasing in second quadrant.
But $\sin 2x$ has higher coefficient, so if $\sin 2x$ is $1$, then $\sin x$ will be $\dfrac{1}{\sqrt{2}}$, but then $3x$ lies in second quadrant which should not be the case.
So $x$ should be at least $\dfrac{\pi}{3}$ so that $3x$ lies in the third quadrant.
So let's check at $x=\dfrac{\pi}{3}$, we get $\sin x$ as $\dfrac{\sqrt{3}}{2}$, $\sin2x$ as $\dfrac{\sqrt{3}}{2}$, $\sin3x$ will be $0$, so if we put these values into the original equation , it would be
$$\dfrac{\sqrt{3}}{2}+\sqrt{3}-0=\frac{3\sqrt{3}}{2}$$
It seems the equation has no solution, is there any better way to solve this problem. I also tried reducing the equation, but it is not getting reduced too much. Is there any better approach to this?
