
Problem 10.1 Find all values of the real parameter $a$ such that the nonnegative solutions of the equation $(2 a-1) \sin x+(2-a) \sin 2 x=$ $\sin 3 x$ form an infinite arithmetic progression.


Solution: Since $\sin 2 x=2 \sin x \cos x$ and $\sin 3 x=\sin x\left(4 \cos ^{2} x-1\right)$, we may write the equation in the form

$\sin x\left(2 \cos ^{2} x-(2-a) \cos x-a\right)=0$. Thus $\sin x=0, \cos x=1$ or $\cos x=-\frac{a}{2}$. The nonnegative solutions of the equations $\sin x=0$ and $\cos x=1$ are $x=k \pi$ and $x=2 k \pi, k=0,1,2, \ldots$ respectively. Let $|a|>2$. In this case the equation $\cos x=-\frac{a}{2}$ has no solution and therefore the nonnegative solutions of the initial equation are $0, \pi, 2 \pi, \ldots$, which form an arithmetic progression. Let now $|a| \leq$ 2 and let $x_{0}$ be the only solution of the equation $\cos x=-\frac{a}{2}$ in the interval $[0, \pi]$. In this case the nonnegative solutions of the last equation are $x=x_{0}+2 k \pi$ and $x=2 \pi-x_{0}+2 k \pi$. It is clear now that the nonnegative solutions form an arithmetic progression only when $x_{0}=0, x_{0}=\frac{\pi}{2}$ and $x_{0}=\pi$, so giving $a=-2, a=0$ and $a=2$. The values of $a$ are $a=0$ and $|a|>2$.
