The equation of the normal to the curve sin y | vedclass

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Question

(A) Given the equation of the curve is $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right) \quad (i)$.At $x = 0$,$\sin y = 0$,which implies $y =

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$2x + \sqrt{3}y = 0$

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(A) Given the equation of the curve is $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right) \quad (i)$.At $x = 0$,$\sin y = 0$,which implies $y = 0$.Now,differentiating equation $(i)$ with respect to $x$ using the product rule:$\cos y \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + y\right) + \sqrt{3} x \cos \left(\frac{\pi}{6} + y\right) \frac{dy}{dx}$.At the point $(0, 0)$,substituting $x = 0$ and $y = 0$:$\cos(0) \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + 0\right) + \sqrt{3}(0) \cos \left(\frac{\pi}{6} + 0\right) \frac{dy}{dx}$.$1 \cdot \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6}\right) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$.Thus,the slope of the tangent $m_t = \frac{\sqrt{3}}{2}$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$.The equation of the normal at $(0, 0)$ is given by $y - y_1 = m_n(x - x_1)$:$y - 0 = -\frac{2}{\sqrt{3}}(x - 0)$.$\sqrt{3}y = -2x$,which simplifies to $2x + \sqrt{3}y = 0$.

The equation of the normal to the curve $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right)$ at $x = 0$ is:

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