Show that $\sin(x)-x\cos(x)=1$ has just a real root in $[\frac{-\pi}{2},\frac{\pi}{2}]$.
We also have $$f:[\frac{-\pi}{2},\frac{\pi}{2}] \rightarrow \mathbb R, f(x)=\frac{x}{\sin^2x}$$
and
$$g: [\frac{-\pi}{2},\frac{\pi}{2}] \rightarrow \mathbb R, f(x)=xf(x)$$
This is how I would go about it:
I have no clue how to solve $\sin(x)-x\cos(x)=0$ so I think maybe it has something to do with $f$ or $g$.
$$\sin(x)-x\cos(x)=0$$
$$\sin(x)=x\cos(x)$$
I raise both side to the second power:
$$\sin^2(x)=x^2\cos^2(x)$$
$$\frac{x^2}{\sin^2(x)}=\frac{1}{\cos^2(x)}$$
The first term is g(x) which means (if I have this right?) that I have to show that the graph of $g(x)$ is tangent to $\frac{1}{\cos^2(x)}$ for any $x\in[\frac{-\pi}{2},\frac{\pi}{2}]$
What do I do now? I appreciate any help!
An idea: define $\;f(x)=\sin x-x\cos x-1\;$ . This is clearly a differentiable function on $\;\left(-\cfrac\pi2,\,\cfrac\pi2\right)\;$ and:
$$f'(x)=\cos x-\cos x+x\sin x=x\sin x\ge 0\,,\,\,\forall x\in\left(-\frac\pi2,\,\frac\pi2\right)\implies $$
the function is injective ($\;1-1\;$) and increasing . But $\;f(-\pi/2)=-2\;,\;\;f(\pi/2)=0\;$ so this means the equation $\;f(x)=0\;$ has no soltuion at all in the given open interval (but if it had then it'd be unique...). Could it be you were given another interval?