# doubt in Solving an inequality

I took upon following question-

Find the values of $$m$$ for which given equation has real roots

$$\sin^2 x-(m-3)\sin x+m=0$$

So I started by first satisfying $$D>=0$$ and found that $$m$$ should range from $$(-\infty,1] \cup [9,\infty)$$

However after this, I found roots by quadratic formula as

$$\sin x = (m-3)/2 \pm \sqrt{(m-1)(m-9)/4}$$

However this value should be within range of sin function and I'm unable to solve the inequality so formed for values of $$m$$.

How to solve the inequality( actually the $$\pm$$ term is confusing me)?

• What do you denote $D$? Commented May 22, 2021 at 9:23
• @Bernard Presumably the discriminant Commented May 22, 2021 at 9:28
• I think it means $b^2-4ac$ from the quadratic rules
– 00GB
Commented May 22, 2021 at 9:29
• $sin$ is supposed to be within the range $[-1, 1]$, so solve the inequality for that too. Commented May 22, 2021 at 9:30
• @ShubhamJohri: I guessed so, but I pointed that when one uses a non-standard notation, it has to be explained. Commented May 22, 2021 at 9:32

When $$m=1$$, both roots of the quadratic are $$-1$$ so a solution for $$x$$ exists. When $$m\in(-\infty,1)$$, the smaller root $$r_1=(m-3)/2-\frac12\sqrt{(m-1)(m-9)}<-1$$ as $$(m-3)/2<-1$$. So we should enforce that the larger root $$r_2=(m-3)/2+\frac12\sqrt{(m-1)(m-9)}\in[-1,1]$$, i.e.\begin{align*}&-1\le(m-3)/2+\frac12\sqrt{(1-m)(9-m)}\le1\\& \iff1-m\le\sqrt{(1-m)(9-m)}\le5-m\end{align*}Note that the first inequality is always true since $$9-m\ge1-m\ge0$$ and hence $$(1-m)(9-m)\ge(1-m)^2$$. The second inequality on squaring gives$$m^2-10m+9\le m^2+25-10m$$which is true. So all values of $$m\le1$$ will work.
When $$m\ge9$$, the larger root $$r_2\ge3$$. So we must enforce $$-1\le r_1\le 1$$, i.e. \begin{align*}&-1\le(m-3)/2-\frac12\sqrt{(m-1)(m-9)}\le1\\& \iff1-m\le-\sqrt{(m-1)(m-9)}\le5-m\\& \iff m-5\le\sqrt{(m-1)(m-9)}\le m-1 \end{align*} Correspondingly note that the second inequality is always true this time around. Squaring the first inequality would yield $$25\le9$$, which is never true. So no $$m\ge9$$ works.
Our final answer is $$m\le1.~\blacksquare$$