2 Roots and Sine!

Consider $g(x)=x^3-(2a)x^2+(b-a)x+(ab)$ . If $f(x)=0$ , then $g(x)=0$ . If $g(x)=0$ , there can be either $1$ solution or $3$ solutions. However, since $f(x)$ has $2$ distinct roots, then $(x-2)$ needs to be a factor of $g(x)$ . By the Remainder Factor theorem, $2^3-2a(2^2)+(b-a)(2)+(ab)=0$ which simplifies to $ab-10a+2b+8=0$ .

Since $f(\sin(x))=0$ has $1$ solution in the given interval, then $\sin(x)$ can only take $1$ distinct value in that interval. Since $\sin(1.5π)=-1$ is the only value that satisfies the condition, and $-1≤\sin(x)≤1$ so even though $(\sin(x)-2)$ is a factor, $\sin(x)=2$ does not bring any real roots. As a result, $(x+1)$ is also a factor of $g(x)$ . Again, by the Remainder Factor Theorem, $(-1)^3-2a(-1)^2+(b-a)(-1)+(ab)=0$ which simplifies to $ab-a-b-1=0$ .

By solving the system regarding $a$ and $b$ by setting both equations to b and solving by substitution and the Quadratic Formula, $a=2$ and $b=3$ . This means that $g(x)=x^3-4x^2+x+6$ . By Polynomial division, the quotient of $g(x)$ and $(x-2)(x+1)$ is $(x-3)$ . As a result, $(x-3)$ is the remaining factor of $g(x)$ . This means that $(x-3)$ is also a factor of $f(x)$ , making $x=3$ and $x=-1$ the $2$ distinct roots. Finally, $-1*3=-3$ which is the final answer.