Find number of roots of $x^2 - x\sin x -\cos^2 x =0 $
In original IIT problem it was $\cos x$ instead of $\cos^2 x$ and then it is pretty easy.
you have to find number of roots and not prove that there exists root.
The answer is 2 for both questions by graphing tool
Hint: first let $f(x) = x^2 - x\sin x -\cos^2 x$; note that $f(x)$ is continuous $\forall x \in \Bbb{R}$,$f(0)=0-0-(1)^2=-1\lt0$ and $f(\pi)=\pi^2-0-(-1)^2=\pi^2-1\gt0 \Rightarrow$ for Weiestrass' Theorem there exists at least one point $c$ $\in [0,\pi]$ such that $f(c)=0$, so there is at least one root $\in [0,\pi]$. Now, make the same reasoning in the interval $[-\pi,0]$, and you'll get that there is at least one root $\in [-\pi,0]$...and now consider the derivative $f'(x)=2x-\sin x-x\cos x+2\cos x \sin x=x(2-\cos x)+\sin 2x-\sin x$. If you look at the expression of the derivative function, you'll note that $f'(x) \gt 0 \forall x\gt 0\Rightarrow$ there exists only one root $\gt 0$. Similarly when $x \lt 0$...so the total number of roots is $2$.