If f(x)=0 is a quadratic equation such that f(−π)=f(π)=0 and fπ2=−3π24, then limx→−π f(x)sin⁡(sin⁡x), is equal to

It is given that f(−π)=f(π)=0. Therefore, -π and π are zeros of f(x).∴    f(x)=a(x+π)(x−π),a≠0⇒    fπ2=−3aπ24⇒−3π24=−3aπ24⇒a=1∴    f(x)=(x+π)(x−π)So, limx→−π f(x)sin⁡(sin⁡x)=limx→−π (x−π)(x+π)sin⁡(sin⁡x)sin⁡x−xsin⁡x=−limx→−π (x−π)sin⁡(sin⁡x)sin⁡x×(x+π)sin⁡(π+x)=−(−π−π)1×1=2π