The number of points in the rectangle : $\{(x,y)|-10\le x\le10$ and $-3\le y\le3\}$ which lie on the curve $y^2=x+\sin x$ and at which the tangent to the curve is parallel to the $X-$axis, is
A) $0$
B) $1$
C) $2$
D) $4$
I am not allowed to use any graphing device or calculator. I had used the fact that derivative of y wrt x should be $0$. From that, I had concluded that $\cos x=-1$. I am unable to do anything after that.
Differentiating implicitly, $$2y\,\frac{dy}{dx}=1+\cos x\ .$$ If $dy/dx=0$ then $\cos x=-1$, so $x=(2n+1)\pi$ for some integer $n$. Since $-10\le x\le10$ the possibilities are $$x=-3\pi,\ {-\pi},\ \pi,\ 3\pi\ .$$ These all give $\sin x=0$ and so $y^2=x$; since $y^2\ge0$ the first two $x$ values are ruled out. This leaves $y^2=\pi$ or $y^2=3\pi$; however since $-3\le y\le3$ we have $y^2\le9$ and the last option is ruled out. Thus $$y=\pm\sqrt\pi$$ and there are two points.
Comment. An extra consideration which is not needed in this question, but might be in others. If one of our values came out as $y=0$, then our very first equation would not guarantee that $\frac{dy}{dx}=0$. We would have to very carefully check $$\lim_{x\to x_0}\frac{1+\cos x}{2y}$$ and see whether or not we get $0$.