To solve the given problem, we need to determine the number of points where the function \(f(x) = [\sin x + \cos x]\) is not continuous in the interval \((0, 2\pi)\).
First, let's understand the expression \(f(x) = [\sin x + \cos x]\). Here, \([\cdot]\) denotes the greatest integer (floor) function, which maps a real number to the largest integer less than or equal to it. Therefore, the function \(f(x)\) is not continuous at precisely those points where \(\sin x + \cos x\) is an integer.
The function \(\sin x + \cos x\) can be re-written using the trigonometric identity:
\(A\sin(x+\phi) = \sin x + \cos x\)
where \(A = \sqrt{2}\) and \(\phi = \frac{\pi}{4}\).
This gives:\( \sin x + \cos x = \sqrt{2} \sin(x+\frac{\pi}{4}) \). The range of \(\sqrt{2} \sin(x + \pi/4)\) is \([- \sqrt{2}, \sqrt{2}]\).
We are interested in the points where \(\sin x + \cos x\) is an integer. Within \([- \sqrt{2}, \sqrt{2}]\), the possible integer values are \(-1, 0,\) and \(1\).
To find the points where \(\sin x + \cos x\) takes these integer values, we solve for:
The expression \( \sin x + \cos x = -1 \) is equivalent to \( \sqrt{2}\sin(x+\frac{\pi}{4}) = -1 \), which gives \(\sin(x+\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\).
Within one period \((0, 2\pi)\), this equation is satisfied at two points.
Similarly, \( \sin x + \cos x = 0 \equiv \sqrt{2}\sin(x+\frac{\pi}{4}) = 0 \) is satisfied at two distinct points within \((0, 2\pi)\).
Finally, \(\sin x + \cos x = 1\) gives \( \sqrt{2}\sin(x+\frac{\pi}{4}) = 1 \), providing one solution in \((0, 2\pi)\).
So, there are altogether 5 points within \((0, 2\pi)\) where \(f(x)\) is not continuous, corresponding to the discontinuity points of the greatest integer function.
Thus, the correct answer is 5.