Answer
$$y=-x+\pi/2$$
Work Step by Step
First we will find the derivative:
$$y'=(\sin^2x\cos x)'=(\sin^2x)'\cos x+\sin^2x(\cos x)'=
2\sin x(\sin x)'\cos x+\sin^2x(-\sin x)=2\sin x\cos x\cdot \cos x-\sin^3x=\sin x(2\cos^2x-\sin^2x)=\sin x(\cos^2x+\cos 2x)$$The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula:
$$y-y_{0}=y'(x_{0})(x-x_{0}).$$
For $(x_{0},y_{0})=(\pi/2,0)$ we get:
$$y-0=\left.\sin x_{0}(\cos^2x_{0}+\cos2x_{0})\right|_{x_{0}=\pi/2}(x-\pi/2)\Rightarrow y=\sin\pi/2(\cos^2\pi/2+\cos2\pi/2)(x-\pi/2)\Rightarrow y=1\cdot(0+(-1))(x-\pi/2)\Rightarrow y=-x+\pi/2$$
