The equation of the normal to the curve $y = 2 sin x$ at (0, 0) is :

$x+\frac{1}{2}y = 0 $ $x-2y = 0 $ $x-\frac{1}{2}y = 0 $ $x+2y = 0 $

$x+2y = 0 $

The correct answer is Option (4) → $x+2y = 0$ $y = 2 \sin x$ at (0, 0) $\frac{dy}{dx}=2\cos x⇒\left.\frac{dy}{dx}\right]_{x=0}=2$ so slope of normal = $-\frac{1}{2}$ eq → $y=-\frac{1}{2}x⇒x+2y=0$