interesting problem:
let postive integer $m$ such $b>1+\dfrac{1}{m}$ ,show that this equation $$\sin{(m+1)x}+b\sin{(mx)}=0$$ has $m-1$ distinct real roots on $(0,\pi)$
I have solve the simpler case $m=1,2$
m=1 $$\sin{(2x)}+b\sin{x}=0\Longrightarrow \sin{x}(2\cos{x}+b)=0$$ and $b>1+\dfrac{1}{m}=2$,if$x\in(0,\pi)$,then we have $$\sin{x}(b+2\cos{x})>0$$ so for this case no solution
m=2
since $$\sin{3x}+b\sin{2x}=3\sin{x}-4\sin^3{x}+2b\sin{x}\cos{x}= \sin{x}(4\cos^2{x}+2b\cos{x}-1)$$ and by condition we have $b>\frac{3}{2}$,it Just to prove that $$4\cos^2{x}+2b\cos{x}-1=0$$ has only one real solution $x\in(0,\pi)$ when $b>\dfrac{3}{2}$F or a quadratic equation is not hard to get
How to prove $m\ge 3$?
Let $f(x) = \sin((m+1)x)+b\sin(mx)$. Notice that if $x\in(0,\pi)$ satisfied $\cos(mx) = 0$ (i.e. $x = \left(i+\frac{1}{2}\right)\frac{\pi}{m}$ for some $0<i<m$), then $|\sin(mx)| = 1$, and hence $$|f(x)| = |\sin((m+1)x)+b\sin(mx)|\ge |b||\sin(mx)| - |\sin((m+1)x)| >\left(1+\frac{1}{m}\right)(1) - 1 > 0. $$ Hence, if $x\in(0,\pi)$ is a root of $\cos(mx)$, then $x$ is not a root of $f$. So it suffices to find the roots of $g(x) = f(x)/\cos(mx)$ in $(0,\pi)\backslash\{\frac{\pi}{2m},\frac{3\pi}{2m},\dots,\frac{(2m-1)\pi}{2m}\}$. Notice that \begin{align} g(x) &= \frac{\sin((m+1)x)+b\sin(mx)}{\cos(mx)} \\ &= \frac{\sin(mx)\cos(x) + \cos(mx)\sin(x) + b\sin(mx)}{\cos(mx)} \\ &= (b+\cos(x))\tan(mx) + \sin(x). \end{align} Since $b+\cos(x)>0$ for all $x$ (as $b>1$), it suffices to find the roots of $h(x) = g(x)/(b+\cos(x))$, i.e. $$ h(x) = \tan(mx) + \frac{\sin(x)}{b+\cos(x)}. $$ Notice that \begin{align} h'(x) &= m\sec^2(mx) + \frac{\cos(x)}{b+\cos(x)} + \frac{\sin^2(x)}{(b+\cos(x))^2} \\ &= m\sec^2(mx) + \frac{b\cos(x) + \cos^2(x)+\sin^2(x)}{(b+\cos(x))^2} \\ &= m\sec^2(mx) + \frac{b\cos(x)+1}{(b+\cos(x))^2}. \end{align} I now claim that $h'(x)>0$ for all $x\in(0,\pi)$ where $h$ is defined. Notice that this is clearly true when $b\cos(x)+1\ge 0$. For $x$ such that $b\cos(x)+1<0$, we see that $\frac{b\cos(x)+1}{(b+\cos(x))^2}$ is minimized (i.e. attains the greatest negative value) when $b\cos(x)+1$ is minimized and $(b+\cos(x))^2$ is minimized. Since $b>1$, both of these are minimized when $\cos(x) = -1$, i.e. $$\frac{b\cos(x)+1}{(b+\cos(x))^2} \ge \frac{-b+1}{(b-1)^2} = \frac{1}{1-b} > \frac{1}{1-(1+1/m)} = -m.$$ for all $x\in(0,\pi)$. Hence $$ h'(x) > m\sec^2(mx) - m = m\tan^2(mx)\ge 0 $$ for all $x\in(0,\pi)$. Now, for each interval of the form $\left(\left(i-\frac{1}{2}\right)\frac{\pi}{m},\left(i+\frac{1}{2}\right)\frac{\pi}{m}\right)$, with $1\le i\le m-1$, notice that $h$ is strictly increasing on this interval, while \begin{align} \lim\limits_{x\rightarrow \left(\left(i-\frac{1}{2}\right)\frac{\pi}{m}\right)^{+}}{h(x)} &= -\infty \\ \lim\limits_{x\rightarrow \left(\left(i+\frac{1}{2}\right)\frac{\pi}{m}\right)^{-}}{h(x)} &= \infty. \end{align} Hence $h$ has precisely one root in $\left(\left(i-\frac{1}{2}\right)\frac{\pi}{m},\left(i+\frac{1}{2}\right)\frac{\pi}{m}\right)$ for each $1\le i\le m-1$. Now \begin{align} \lim\limits_{x\rightarrow 0^{+}}{h(x)} &= 0\\ \lim\limits_{x\rightarrow\pi^{-}}{h(x)} &= 0 \end{align} so $h$ does not have a root in $\left(0,\frac{\pi}{2m}\right)$ or $\left(\frac{(2m-1)\pi}{2m},\pi\right)$, as it is strictly increasing in these intervals as well. Thus, $h$ has one root in each interval $\left(\left(i-\frac{1}{2}\right)\frac{\pi}{m},\left(i+\frac{1}{2}\right)\frac{\pi}{m}\right)$ for $1\le i\le m-1$ and no roots elsewhere, so $h$ has $m-1$ roots. It follows that $g$, and hence $f$, has $m-1$ roots as well.