Spring Mathematical Tournament

a) We have $$\begin{array}{rl}f(x+\pi )& =a(|\sin (x+\pi )|+|\cos (x+\pi )|)-3\sin (2x+2\pi )-7\\ & =a(|-\sin x|+|-\cos x|)-3\sin (2x)-7=f(x),\\ & \\ f\left(\frac{\pi }{2}-x\right)& =a\left(|\sin \left(\frac{\pi }{2}-x\right)|+|\cos \left(\frac{\pi }{2}-x\right)|\right)-3\sin (\pi -2x)-7\\ & =a\left(|\cos x|+|\sin x|\right)-3\sin (2x)-7=f(x),\\ & \\ f\left(\frac{3\pi }{2}-x\right)& =a\left(|\sin \left(\frac{3\pi }{2}-x\right)|+|\cos \left(\frac{3\pi }{2}-x\right)|\right)-3\sin (3\pi -2x)-7\\ & =a\left(|-\cos x|+|-\sin x|\right)-3\sin (2x)-7=f(x).\end{array}$$

b) Direct verification shows that for every integer $k$ we have $f(\frac{k\pi }{2})=a-7$. Moreover $f(\frac{\pi }{4})=a\sqrt{2}-10$ and $f(\frac{3\pi }{4})=a\sqrt{2}-4$. If $a\ne 7$, $a\ne 5\sqrt{2}$ and $a\ne 2\sqrt{2}$, according to a) the equation $f(x)=0$ has even number of roots in any of the intervals $(0,\frac{\pi }{2})$ $(\frac{\pi }{2},\pi )$ and therefore it has even number of roots in the interval $(0,n\pi )$.

1. If $a=7$ then $f(x)=7(|\sin x|+|\cos x|)-3\sin 2x-7$ and $f(\frac{\pi }{2})=0$.

1.1. Let $x\in (0,\frac{\pi }{2})$. Then $f(x)=7(\sin x+\cos x)-3\sin 2x-7$. Set $y=\sin x+\cos x$. Then $y=\sqrt{2}\sin (x+\frac{\pi }{4})\in (1,\sqrt{2}]$ ($\sin 2x=y^2-1$) and the equation $f(x)=0$ becomes $3y^2-7y+4=0$. Therefore $y_1=1$ and $y_2=\frac{4}{3}$. Hence $y_2=\frac{4}{3}$ and the equation $f(x)=0$ has 2 roots in $(0,\frac{\pi }{2})$.

1.2. Let $x\in (\frac{\pi }{2},\pi )$. Then $f(x)=7(\sin x-\cos x)-3\sin 2x-7$. Set $y=\sin x-\cos x$. Then $y=\sqrt{2}\sin (x+\frac{\pi }{4})\in (1,\sqrt{2}]$ and $f(x)=0$ is equivalent to $3y^2+7y-10=0$. Therefore $y_1=1$ and $y_2=-\frac{10}{3}$, i.e. $f(x)=0$ has no solutions in the given interval. Therefore $f(x)=0$ has 3 roots in the interval $(0,\pi )$. The total number of roots in the interval $(0,n\pi )$ equals $3n+n-1=4n-1$ and $4n-1=2007$ implies $n=502$.

2. If $a=5\sqrt{2}$ then $f(x)=5\sqrt{2}(|\sin x|+|\cos x|)-3\sin 2x-7$.

2.1. Let $x\in (0,\frac{\pi }{2})$. Then $f(x)=5\sqrt{2}(\sin x+\cos x)-3\sin 2x-7$ and setting $y=\sin x+\cos x$ yields that $f(x)=0$ is equivalent to $3y^2-5\sqrt{2}y+4=0$, $y\in (1,\sqrt{2}]$. Therefore $y_1=\sqrt{2}$ and $y_2=\frac{2\sqrt{2}}{3}<1$, i.e. only $x=\frac{\pi }{4}$ is a solution.

2.2. Let $x\in (\frac{\pi }{2},\pi )$. Then $f(x)=5\sqrt{2}(\sin x-\cos x)-3\sin 2x-7$ and setting $y=\sin x-\cos x$ yields that $f(x)=0$ is equivalent to $3y^2+5\sqrt{2}y-10=0$, $y\in (1,\sqrt{2}]$. The roots of this equation do not belong to the interval $(1,\sqrt{2}]$. In this case there is 1 root in $(0,\pi )$ and $n$ roots in $(0,n\pi )$. Therefore $n=2007$.

3. When $a=2\sqrt{2}$ analogous observations show that there is a unique root $x=\frac{3\pi }{4}$ in the interval $(0,\pi )$ and $n$ roots in the interval $(0,n\pi )$. Again $n=2007$.

Answer: $a=7$, $n=502$; $a=5\sqrt{2}$, $n=2007$; $a=2\sqrt{2}$, $n=2007$.