If `[sin x]+[sqrt(2) cos x]=-3 , x in [0,2pi]`, (where ,[.] denotes th greatest integer function ), then

Options

A. `x in (pi,(5pi)/4)`
B. `x in (pi, (7pi)/6)`
C. `x in [pi,(5pi)/(4)]`
D. None of these

Answer

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

### Step 1: Understand the ranges of \(\sin x\) and \(\cos x\)
The range of \(\sin x\) is \([-1, 1]\) and the range of \(\cos x\) is also \([-1, 1]\).

### Step 2: Determine the possible values of \([ \sin x ]\)
Since \(\sin x\) can take values from \(-1\) to \(1\):
- \([ \sin x ]\) can be \(-1\) (for \(-1 < \sin x < 0\)), \(0\) (for \(0 \leq \sin x < 1\)), or \(1\) (only when \(\sin x = 1\)).

### Step 3: Determine the possible values of \([ \sqrt{2} \cos x ]\)
The value of \(\sqrt{2} \cos x\) ranges from \(-\sqrt{2}\) to \(\sqrt{2}\) (approximately \(-1.414\) to \(1.414\)):
- \([ \sqrt{2} \cos x ]\) can be \(-2\) (for \(-\sqrt{2} < \sqrt{2} \cos x < -1\)), \(-1\) (for \(-1 < \sqrt{2} \cos x < 0\)), \(0\) (for \(0 \leq \sqrt{2} \cos x < 1\)), or \(1\) (only when \(\sqrt{2} \cos x = 1\)).

### Step 4: Set up the equation
We need to satisfy the equation:
\[ [ \sin x ] + [ \sqrt{2} \cos x ] = -3 \]
The only way to achieve \(-3\) is if:
\[ [ \sin x ] = -1 \quad \text{and} \quad [ \sqrt{2} \cos x ] = -2 \]

### Step 5: Analyze the conditions
1. **Condition for \([ \sin x ] = -1\)**:
- This occurs when \(-1 < \sin x < 0\), which corresponds to the interval \(x \in (\pi, 2\pi)\).

2. **Condition for \([ \sqrt{2} \cos x ] = -2\)**:
- This occurs when \(-2 < \sqrt{2} \cos x < -1\), which simplifies to:
\[ -\sqrt{2} < \cos x < -\frac{1}{\sqrt{2}} \]
- The value \(-\frac{1}{\sqrt{2}}\) corresponds to angles \(x = \frac{3\pi}{4}\) and \(x = \frac{5\pi}{4}\).

### Step 6: Find the overlapping intervals
From the conditions:
- For \([ \sin x ] = -1\): \(x \in (\pi, 2\pi)\)
- For \([ \sqrt{2} \cos x ] = -2\): \(x \in \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right)\)

The overlapping interval is:
\[ x \in \left(\pi, \frac{5\pi}{4}\right) \]

### Step 7: Conclusion
Thus, the solution for the equation \([ \sin x ] + [ \sqrt{2} \cos x ] = -3\) in the interval \([0, 2\pi]\) is:
\[ x \in \left(\pi, \frac{5\pi}{4}\right) \]