$f(x) = g(x)$ for exactly two values of $x$ Let $(f,g):\mathbb R\to \mathbb R$, $f(x) = x^2 - \frac{\cos x}{2}$ and $g(x)= \frac{x\sin x}{2}$
These are the given options :-
(A) $f(x) = g(x)$ for more than two values of $x$
(B) $f(x) \neq g(x)$, for all $x \in \mathbb R$
(C) $f(x) = g(x)$ for exactly one value of $x$
(D) $f(x) = g(x)$ for exactly two values of $x$
I broke down the trigonometric terms into power series centred at $0$ and got this 
$$f(x)= x^2 -[1-\frac{x^2}{2}] = g(x) = \,x^2\\
x^2=1 \Rightarrow x = \pm 1 \;\text{when } x \to 0$$
Isn't it contradicting that we assumed $x$ to be around $0$, which is why we wrote the trig terms in power series, but the answer I got is far away from $0$.
Answer given is $D$ , but how is it possible?
 A: Note that $f(0)=-\frac12<0=g(0)$.
On the other hand, $f'(x)-g'(x)=\frac12x\bigl((4-\cos(x)\bigr)$ and therefore $f-g$ is strictly increasing in $[0,+\infty)$. This, together with fact that  $f(\pi)>g(\pi)$, shows that the equality $f(x)=g(x)$ takes place exatly once in $[0,+\infty)$. And, since both functions are even, that equality takes place exactly twice in $\mathbb R$.
A: Let $$h(x)=f(x)-g(x)=x^2-\frac12x\sin x-\frac12\cos x$$ We know that $h$ is continuous over $\mathbb{R}$ so we find its derivative $$h'(x)=2x-\frac12\sin x-\frac12x\cos x+\frac12\sin x=x\left(2-\frac12\cos x\right)=0$$ for stationary points so either $x=0$ or $\cos x=4$ which is impossible. 
Now $$h''(x)=2-\frac12\cos x+\frac12x\sin x\implies h''(0)=\frac32>0$$ so $0$ is a global minimum.
Since $h(-1)=0.309...$ and $h(1)=0.309...$ by the Intermediate Value Theorem, there are at least two points of intersection.
Also, since $h$ is decreasing in $(-\infty,0)$ and increasing in $(0,\infty)$, we must have that there are exactly two solutions to the equation $h(x)=0\implies f(x)=g(x)$.
