# Least value of $a$ for which at least one solution exists?

What is the least value of $a$ for which $$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$ has atleast one solution in the interval $(0,\frac{\pi}{2})$?

I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points. This gives $$\sin(x)=\frac{2}{3}$$ as $\cos(x)$ is not $0$ in $(0,\frac{\pi}{2})$. I calculate $f''(x)$ and at $\sin(x)=\frac{2}{3}$, I get a minima. Now to have at least one solution, putting $\sin(x)=\frac{2}{3}$ in the main equation, I get $f=9-a$, which should be greater than or equal to $0$. I then get the 'maximum' value of $a$ as $9$. Where did I go wrong? [Note the function is $f(x)=LHS-RHS$ of the main equation.]

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You're almost there. The minimum of the function in the interval is $9$, which is $a$. You are done. – Ron Gordon Feb 19 '13 at 13:15
The question needs the minimum value of $a$. Do you think it can be wrong( it could have asked instead for maximum value)? – Ashish Gaurav Feb 19 '13 at 13:18
No, I think it is OK. – Ron Gordon Feb 19 '13 at 13:19
You can save yourself some writing and maybe some confusion by noting that over $(0,\frac \pi 2), \sin x$ varies over $(0,1)$ so define $y=\sin x$ and work with $y$. – Ross Millikan Feb 19 '13 at 14:10

One possible approach: Find a common denominator, then :

$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a\iff \frac{4(1- \sin x) + \sin x}{\sin x - \sin^x} = a$$ $$\iff 4-3\sin x = a(\sin x - \sin^2 x)\tag{\sin x \neq 0}$$

Now write the equation as a quadratic equation in $\sin x$:

$$a\sin^2 x - (3 + a)\sin x + 4 = 0$$

You can solve for when the equation has a real solution (by determining when the discriminant is greater than or equal to 0). $$b^2 - 4ac \geq 0 \iff (3+a)^2 - 16 a \geq 0 \iff a^2 -10a + 9 \geq 0 \iff (a - 1)(a-9) \geq 0$$

Then determine which values of $a$ satisfy the inequality and give in the desired interval.

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Doing your way, I get the quadratic as $a{sin(x)}^2-(a+3)sin(x)+4=0$, and since $sin(x)$ is real, so $D>=0$ This gives $a>=9$ or $a<=1$, where $a>0$. Any explanation why we reject the latter(I mean it could also be a correct range for $a$, if this were not the question)? – Ashish Gaurav Feb 19 '13 at 13:22
@AshishGaurav Note the left hand side of the equation is at least $4$. – David Mitra Feb 19 '13 at 13:31
Thanks, this worked--! – Ashish Gaurav Feb 19 '13 at 13:32
Great! You're welcome! – amWhy Feb 19 '13 at 13:33
You don't need to solve the equation (and your factorization is incorrect). You need to determine when the quadratic has a real solution. So, determine when the discriminant is nonnegative. – David Mitra Feb 19 '13 at 13:35

My Solution:: Using the Cauchy-Schwarz inequality:: $\displaystyle \frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$

and equality holds when $\displaystyle \frac{a}{x} = \frac{b}{y}.$

So here $\displaystyle \frac{2^2}{\sin x}+\frac{1^2}{1-\sin x}\geq \frac{(2+1)^2}{\sin x+1-\sin x}\Rightarrow a\geq 9$

and equality holds when $\displaystyle \frac{2}{\sin x} = \frac{1}{1-\sin x}\Rightarrow \sin x = \frac{2}{3}$

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