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Problem Find the equation of the normal to the curve y = sin x y=\sin x y = sin x x = π 6 x=\dfrac{\pi}{6} x = 6 π y − 1 2 = 3 2 ( x − π 6 ) y-\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\left(x-\dfrac{\pi}{6}\right) y − 2 1 = 2 3 ( x − 6 π ) y − 1 2 = − 2 3 ( x − π 6 ) y-\dfrac{1}{2}=-\dfrac{2}{\sqrt{3}}\left(x-\dfrac{\pi}{6}\right) y − 2 1 = − 3 2 ( x − 6 π ) y − 1 2 = 2 3 ( x − π 6 ) y-\dfrac{1}{2}=\dfrac{2}{\sqrt{3}}\left(x-\dfrac{\pi}{6}\right) y − 2 1 = 3 2 ( x − 6 π ) y − 1 2 = − 3 2 ( x − π 6 ) y-\dfrac{1}{2}=-\dfrac{\sqrt{3}}{2}\left(x-\dfrac{\pi}{6}\right) y − 2 1 = − 2 3 ( x − 6 π ) Browse the problem library