If `[sinx]+[sqrt(2)cosx]=-3,x in[0,2pi]([.]-GIF), "then" x in`

Options

A. `((5pi)/(4),2pi)`
B. `[(5pi)/(4),2pi]`
C. `(pi,(5pi)/(4))`
D. `[pi,(5pi)/(4)]`

Explanation

To solve the equation \([sin(x)] + [\sqrt{2}cos(x)] = -3\) for \(x\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Understand the Greatest Integer Function (GIF)
The greatest integer function \([y]\) returns the largest integer less than or equal to \(y\). For the equation to equal \(-3\), both \([sin(x)]\) and \([\sqrt{2}cos(x)]\) must be negative integers that sum to \(-3\).

### Step 2: Determine Possible Values for \([sin(x)]\) and \([\sqrt{2}cos(x)]\)
Since \([sin(x)]\) can take values of \(-1\) or \(0\) (as \(sin(x)\) ranges from \(-1\) to \(1\)), we analyze the possible combinations:
1. \([sin(x)] = -1\) and \([\sqrt{2}cos(x)] = -2\)
2. \([sin(x)] = -2\) and \([\sqrt{2}cos(x)] = -1\) (not possible since \([sin(x)]\) cannot be less than \(-1\))

Thus, we only need to consider the first case:
- \([sin(x)] = -1\) implies \(sin(x) \in [-1, 0)\)
- \([\sqrt{2}cos(x)] = -2\) implies \(\sqrt{2}cos(x) \in [-2, -1)\), which means \(cos(x) \in [-\sqrt{2}, -\frac{1}{\sqrt{2}})\)

### Step 3: Analyze the Conditions
1. For \([sin(x)] = -1\):
- This occurs when \(sin(x) = -1\), which happens at \(x = \frac{3\pi}{2}\).

2. For \([\sqrt{2}cos(x)] = -2\):
- This condition means \(cos(x) < -\frac{1}{\sqrt{2}}\), which occurs in the intervals where \(x\) is in the third and fourth quadrants.

### Step 4: Find the Intersection of Conditions
- At \(x = \frac{3\pi}{2}\):
- \(sin\left(\frac{3\pi}{2}\right) = -1\) which gives \([sin(x)] = -1\).
- \(cos\left(\frac{3\pi}{2}\right) = 0\) which gives \([\sqrt{2}cos(x)] = [0] = 0\), which does not satisfy the condition.

### Step 5: Check Other Values
To satisfy both conditions, we need to find \(x\) such that:
- \(sin(x) = -1\) (only at \(x = \frac{3\pi}{2}\))
- \(cos(x)\) must also be in the range \([-1, -\frac{1}{\sqrt{2}})\) which does not hold at \(x = \frac{3\pi}{2}\).

### Conclusion
The only valid solution occurs when:
- \(x = \frac{7\pi}{4}\) (where \(sin(x) = -\frac{\sqrt{2}}{2}\) and \(cos(x) = -\frac{\sqrt{2}}{2}\)).

Thus, the solution set is:
\[ x = \frac{7\pi}{4} \]