The slope of the tangent line to \(y=3\sin^2\dfrac x2\) is
\(y'(x)=3\sin\dfrac x2\cos\dfrac x2=\dfrac32\sin x\implies y'(\pi)=0\)
which means the tangent line at this point is horizontal.
Because the tangent line passes through
\(y(\pi)=3\sin^2\dfrac\pi2=3\)
the tangent line has equation \(y=3\).
