Let $f(x)=[4+3\cos x],\,x∈\left(-\frac{π}{2},\frac{π}{2}\right)$, x where [x] = greatest integer less than or equal to x. The number of points of discontinuity of f (x) is

5

4 < 4 + 3 cos x ≤ 7 for $x∈\left(-\frac{π}{2},\frac{π}{2}\right)$

f (x) = [4 + 3 cos x] is discontinuous at those points where 4 + 3 cos x is an integer.

Now 4 + 3 cos x = 4 if cos x = 0. So $x=\left(\frac{-π}{2},\frac{π}{2}\right)$ (not possible)

4 + 3 cos x = 5 if $\cos x=\frac{1}{3}$. So x has two values $\cos^{-1}(\frac{1}{3})$ and $-\left(\cos^{-1}\frac{1}{3}\right)$

4 + 3 cos x = 6 if $\cos x=\frac{2}{3}$. So x has two values $\cos^{-1}(\frac{2}{3}),\cos^{-1}(\frac{2}{3})$

4 + 3 cos x = 7 if cos x =1. So x = 0

∴ the number of values of x = 2 + 2 + 1 = 5