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The Question

If the equation $sin^2x -k\space sinx -3 =0$ has exactly two distinct real roots in $[0,\pi]$, then find the values of $k$.

The Answer

Let $f(t)=t^2 -kt -3$, where $t=sinx$.
Since equation has exactly two distinct real roots in $[0,\pi]$, $f(t)=0$ must have exactly one root in $(0,1)$.
Now, $f(0)=-3$.
So, we must have $f(1)=-k-2>0$
or $k<-2$

My Question

Why does $f(t)=0$ have exactly one root in $(0,1)$ ? Why not $2$ ? I guess this might be a trivial question but I really don't get it.

UNAN
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Anili
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    Unless $t=1$, the equation $t=\sin x$ has two solutions in $[0,\pi]$. Thus, you need a unique solution for $t$ in $[0,1)$ to have two solutions for $x$. Since $t=0$ is never a solution, it is enough to consider $t\in(0,1)$. – anankElpis Jul 05 '21 at 15:03
  • Oh that makes sense. Thank you so much! – Anili Jul 05 '21 at 15:41

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