Trigonometric Equation = Quadratic Equation???

We can rewrite $f(x)$ as $f(x)=\cos{(x-A)}*\sqrt{(a^2+b^2)}$ where $A=\arctan{(a/b)}$ . Thus, $\cos{(x-A)}=$ $\frac{ay^2+by}{\sqrt{(a^2+b^2)}}$ if $f(x)=g(y)$ . In the given interval, there is only one value of $x$ if $\cos{(x-A)}=1$ , so $\tan{x}=$ $\frac{a}{b}$ . Again, since $x$ can only take $1$ value, $\tan{x}=0$ . This means that $a=0$ , and $by=\sqrt{b^2}$ . Solving this equation yields $y=1$ , and $4|y|-3=1$ which is the final answer.