Let $f(x)=f\left(x\right)=\frac{x}{\sin x}$ & $x∈\left(0,\frac{\pi}{2}\right)$, Then prove that the interval in which at least one root of equation $h\left(x\right)$= $\frac{2}{x-f\left(\frac{\pi}{12}\right)}+\frac{3}{x-f\left(\frac{\pi}{4}\right)}+\frac{4}{x-f\left(\frac{5\pi}{12}\right)}$ lie is:
$\left(f\left(\frac{\pi}{12}\right),f\left(\frac{\pi}{4}\right)\right)$ and $\left(f\left(\frac{\pi}{4}\right),f\left(\frac{5\pi}{12}\right)\right)$
My Approach: I tried to apply rolle's theorem here and made a function $g\left(x\right)=2\ln\left(x-f\left(\frac{\pi}{12}\right)\right)+3\ln\left(x-f\left(\frac{\pi}{4}\right)\right)+4\ln\left(x-f\left(\frac{5\pi}{12}\right)\right)$ and tried to find the interval in which the values of g(x) become equal and then since h(x) is the derivative of g(x), I could prove that h(x) has at least one root in that interval. But I don't think that helps. Any hint on how to proceed further is appreciated.
I don't even know if I should use rolle's theorem here or not.
First note that $f'(x)=\frac {\sin x -x\cos x} {\sin^{2}x} >0$ on $(0,\frac {\pi} 2)$ (because $\tan x >x$ in that interval). This proves that $f$ is incearsing so $f(\frac {\pi} {12}) <f(\frac {\pi} {4})<f(\frac {5\pi} {12})$.
So $h(x) >0$ for $x >\frac {5\pi} {12}$ and $h(x) <0$ for $x <\frac {\pi} {12}$. It remains to prove that there is at least one zero in the two intervals given. For the first interval the left hand-limit at the right end point ($\lim_{x \to \frac {\pi} 4-} h(x)$) is $-\infty$ and the right-hand limit at the left end point is $+\infty$. By IVP of continuous functions $h$ has at least one zero in the interval. A silmilar argument holds for the second interval.