If the equation $\sin^2x-a\sin x+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?

Question

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?

Answer 1

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 $$ b\ge 1\ \ \ \text{ or }\ \ \ b\le 0. $$

Answer 2

Hint; Since we have $$|\sin(x)|\le 1$$ you have to solve $$\left|\frac{1}{2}\left(a\pm\sqrt{a^2-4b}\right)\right|\le 1$$

Answer 3

To $x$ to have only one solution in the range $(0,\pi)$, then $x$ must be $\pi\over 2$, according to the unit circle for trigonometry.
Since, $\sin{\frac{\pi}{2}} = 1$, $$\sin^2 x - a\sin x +b = 1-a+b =0$$ $$\implies a-b=1 \to\text{eq.1} $$ From what you have tried we get, $$a^2 =4b \to \text{eq.2} $$ Thus, to satisfy eqs. 1&2 is, From eq.1 , we get, $$a=1+b$$ Subtituting into eq.2, $$(1+b)^2=4b$$ $$1+2b+b^2=4b$$ $$(b-1)^2=0$$ $$\therefore b=1$$ and $a=2$