IF `[5 sin x]+ [cos x]+6=0,` then range of `f(x)=sqrt3 cos x + sin ` x corresponding to solution set of the given equation is: (where `[.]` denotes greatest integer function)

Options

A. `[-2,-1]`
B. `(-(3sqrt3+2)/(5),-1)`
C. `[-2,-sqrt3)`
D. `(-(3sqrt3+4)/(5), -1)`

Answer

To solve the equation \([5 \sin x] + [\cos x] + 6 = 0\) and find the range of \(f(x) = \sqrt{3} \cos x + \sin x\) corresponding to the solution set of the given equation, we will follow these steps:

### Step 1: Analyze the given equation
The equation is given as:
\[ [5 \sin x] + [\cos x] + 6 = 0 \]
This can be rewritten as:
\[ [5 \sin x] + [\cos x] = -6 \]

### Step 2: Determine possible values of \([5 \sin x]\) and \([\cos x]\)
We know that:
- \(\sin x\) ranges from \(-1\) to \(1\), hence \(5 \sin x\) ranges from \(-5\) to \(5\).
- The greatest integer function \([5 \sin x]\) can take values from \(-5\) to \(5\), specifically \(-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\).

For \(\cos x\), which also ranges from \(-1\) to \(1\):
- The possible values of \([\cos x]\) are \(-1, 0, 1\).

### Step 3: Find combinations that satisfy the equation
We need to find combinations of \([5 \sin x]\) and \([\cos x]\) such that:
\[ [5 \sin x] + [\cos x] = -6 \]
The only combination that satisfies this is:
\[ [5 \sin x] = -5 \quad \text{and} \quad [\cos x] = -1 \]

### Step 4: Determine the intervals for \(\sin x\) and \(\cos x\)
From \([5 \sin x] = -5\):
\[ -5 \leq 5 \sin x < -4 \implies -1 \leq \sin x < -\frac{4}{5} \]

From \([\cos x] = -1\):
\[ -1 \leq \cos x < 0 \]

### Step 5: Identify the intervals for \(x\)
The intervals for \(\sin x\) and \(\cos x\) give us:
- For \(-1 \leq \sin x < -\frac{4}{5}\), this occurs in the third and fourth quadrants.
- For \(-1 \leq \cos x < 0\), this occurs in the second and third quadrants.

### Step 6: Find the corresponding angles
To find the angles where \(\sin x\) is between \(-1\) and \(-\frac{4}{5}\):
- The angle corresponding to \(\sin x = -\frac{4}{5}\) is approximately \(233^\circ\) to \(307^\circ\) (in the third quadrant).

### Step 7: Determine the range of \(f(x)\)
Now we need to calculate the range of \(f(x) = \sqrt{3} \cos x + \sin x\) for the angles \(x\) in the interval \( (233^\circ, 307^\circ) \).

1. **Minimum value of \(f(x)\)**:
- At \(\sin x = -\frac{4}{5}\) and \(\cos x = -\frac{3}{5}\):
\[ f(x) = \sqrt{3} \left(-\frac{3}{5}\right) + \left(-\frac{4}{5}\right) = -\frac{3\sqrt{3}}{5} - \frac{4}{5} = -\frac{3\sqrt{3} + 4}{5} \]

2. **Maximum value of \(f(x)\)**:
- At \(\sin x = -1\) and \(\cos x = 0\):
\[ f(x) = \sqrt{3}(0) + (-1) = -1 \]

### Final Result
The range of \(f(x)\) corresponding to the solution set of the given equation is:
\[ \left[-\frac{3\sqrt{3} + 4}{5}, -1\right] \]