root of the equation $4\sin ^2x +10 x^2=\cos x$ The equation $4\sin ^2x +10 x^2=\cos x$ has 
1) no real solution.
2) exactly one real solution.
3) exactly two real solution.
4) more than two real solution.
I find $f(0)f(\pi /2)<0$, so in $(0, \pi /2)$ there is a root. But what should be the right option? Please help. Thanks
 A: Let $f(x)=4\sin^2x+10x^2-\cos{x}.$
Thus, $$f''(x)=20+\cos{x}+8\cos2x>0,$$ which says that $f$ is a convex function and
our equation has two roots maximum.
But $f(0)<0$ and $f(1)>0$, which says that our equation has two real roots exactly.
A: Define $f(x)=4\sin^2x+10x^2-\cos x$.
Then $f(x)=f(-x)$, so it suffices to look at $(0,\infty)$.
We have 
$f'(x)=20x+\sin x+4\sin(2x)$
$f''(x)=12+\cos x+16\cos^2x$
$f''$ is clearly positive, so $f'$ is increasing, so $f'(x)>f'(0)=0$, so $f$ is increasing on $(0,\infty)$.
Since there is a root in $(0,\infty)$ (what you already know), and the function increases it has to be the only one. And by $f(x)=f(-x)$ there is also only one in $(-\infty,0)$.
$f(0)\neq 0$, so 0 is not a root.
Two roots.
A: Both the LHS and RHS are even functions, so that the number of roots is even ($x=0$ is not a root).
Then the extrema are given by the roots of
$$8\sin x\cos x+20x=-\sin x.$$
Clearly, $x=0$ is one, and if we write
$$8\text{ sinc } x\cos x+20=-\text{ sinc } x$$ (where $\text{sinc }x:=\dfrac{\sin x}x\le1$), there cannot be any other.
As $\text{LHS}(0)<\text{RHS}(0)$ and $\text{RHS}(x)$ can be negative, there are exactly two roots.
