If the equation $\sin^2(x)-a\sin(x)+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?
What I try:
$$\displaystyle \sin x=\frac{a\pm \sqrt{a^2-4b}}{2}\in\bigg(0,1\bigg)$$
for one real solution $a^2=4b$.
How do I solve it?
[Math] If the equation $\sin^2x-a\sin x+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$
quadraticstrigonometry
Best Answer
If $x\in (0,\pi)$ solves the equation then $\pi-x\in (0,\pi)$ also solves the equation since it holds $\sin x = \sin(\pi-x)$. By the uniqueness of solution on $(0,\pi)$, we have $x=\pi-x$, i.e. $x=\frac{\pi}2$. Hence $\sin(\frac {\pi}2)=1$ solves the quadratic equation, which implies that $$ \sin^2(x)-a\sin x+b=(\sin x-1)(\sin x-b). $$ In order that $\sin x = b$ has no solution on $(0,\pi)$ other than $x=\frac{\pi}2$, it must be that $$ b\ge 1\ \ \ \text{ or }\ \ \ b\le 0. $$