Condition for three roots The equation $$2\sin ^3 x +(2\lambda -3)\sin ^2 x -(3\lambda +2)\sin x -2\lambda=0$$ has excatly three roots in $(0,2\pi)$ then what can be the value of $\lambda$ .
I thought for three roots the differentiation of the equation should have two roots . but from that I am not getting anya graph .

The equation can be factored as follows:
$$
(\sin x - 2)(2\sin x + 1)(\sin x + \lambda) = 0
$$
For the lefthand side to be $0$, one of the factors needs to be $0$.
$\sin x - 2$ cannot be $0$.
$2\sin x + 1 = 0$ has two solutions ($\frac{7\pi}{6}$ and $\frac{5\pi}{3}$). The number of solutions of $\sin x + \lambda = 0$ depends on $\lambda$.
Because the total number of solutions needs to be exactly $3$, we see that $\sin x + \lambda = 0$  has either exactly one solution, this happens when $\lambda$ is $-1$, $0$ or $1$, or there have to be $2$ solutions, and exactly one of these solutions coincides with $\frac{7\pi}{6}$ or $\frac{5\pi}{3}$. This last option is not possible, because then $\lambda$ would be $\frac{1}{2}$, which means that both solutions of $\sin x + \lambda = 0$ are $\frac{7\pi}{6}$ and $\frac{5\pi}{3}$, so the total number of solutions would be 2.
So we see that the only possibilities for $\lambda$ are $-1$, $0$ and $1$.
