The equation of the normal to the curve y = x + sin x cos x at x = $\pi$/2 is ___________ .

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2x = $\pi$ \text { Given }: y = x + \sin x \cos x \text { On differentiating both sides w.r.t.x, we get } dy/dx = 1 + \cos^2 x - \sin^2 x \text { Slope of the tangent }= \left( dy/dx \right)_{x = $\pi$/2} {=1+cos}^2 $\pi$/2 {-sin}^2 $\pi$/2=1-1=0 \text { Slope of the normal, } m=- $\frac{1}{0}$ \text { When }x=$\pi$/2, y=$\pi$/2+cos$\pi$/2\sin$\pi$/2=$\pi$/2 \text { Now }, \left( x_1 , y_1 \right) = \left( $\pi$/2, $\pi$/2 \right) \therefore \text { Equation of the normal } = y - y_1 = m\left( x - x_1 \right) \Rightarrow y - $\pi$/2 = - $\frac{1}{0}$\left( x - $\pi$/2 \right) \Rightarrow x = $\pi$/2 \Rightarrow 2x = $\pi$

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