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

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(B) Given the curve $\sin y = x \sin \left( \frac{\pi}{3} + y \right)$.At $x = 0$,$\sin y = 0 \implies y = 0$. So the point is $(0, 0)$.Differentiatin

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

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(B) Given the curve $\sin y = x \sin \left( \frac{\pi}{3} + y \right)$.At $x = 0$,$\sin y = 0 \implies y = 0$. So the point is $(0, 0)$.Differentiating both sides with respect to $x$:$\cos y \frac{dy}{dx} = \sin \left( \frac{\pi}{3} + y \right) + x \cos \left( \frac{\pi}{3} + y \right) \frac{dy}{dx}$.At $(0, 0)$:$\cos(0) \frac{dy}{dx} = \sin \left( \frac{\pi}{3} \right) + 0 \implies \frac{dy}{dx} = \frac{\sqrt{3}}{2}$.The slope of the normal is $m_n = -\frac{1}{dy/dx} = -\frac{2}{\sqrt{3}}$.The equation of the normal at $(0, 0)$ is $y - 0 = -\frac{2}{\sqrt{3}}(x - 0)$.$\sqrt{3}y = -2x \implies 2x + \sqrt{3}y = 0$.

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

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