Number of solution of `sinx cosx-3cosx+4sinx-13 gt 0` in `[0,2pi]` is equal to

Options

A. `0`
B. `1`
C. `2`
D. `4`

Answer

To solve the inequality \( \sin x \cos x - 3 \cos x + 4 \sin x - 13 > 0 \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ \sin x \cos x - 3 \cos x + 4 \sin x - 13 > 0 \]
We can rearrange this to:
\[ \sin x \cos x + 4 \sin x - 3 \cos x - 13 > 0 \]

### Step 2: Substitute for \(\sin x \cos x\)
Using the double angle identity, we know:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Thus, we can rewrite the inequality as:
\[ \frac{1}{2} \sin 2x + 4 \sin x - 3 \cos x - 13 > 0 \]

### Step 3: Multiply through by 2
To eliminate the fraction, we multiply the entire inequality by 2 (note that this does not change the direction of the inequality):
\[ \sin 2x + 8 \sin x - 6 \cos x - 26 > 0 \]

### Step 4: Analyze the left-hand side
Let \( y = 4 \sin x - 3 \cos x \). The maximum and minimum values of \( y \) can be found using the formula for the maximum and minimum of \( A \sin x + B \cos x \):
\[ \text{Max} = \sqrt{A^2 + B^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
\[ \text{Min} = -\sqrt{A^2 + B^2} = -5 \]

### Step 5: Determine the range of the left-hand side
Thus, the expression \( 4 \sin x - 3 \cos x \) varies between -5 and 5. Therefore:
\[ -5 \leq 4 \sin x - 3 \cos x \leq 5 \]

### Step 6: Compare with the right-hand side
Now, we need to analyze the right-hand side:
\[ \sin 2x = 2 \sin x \cos x \]
The maximum value of \( \sin 2x \) is 1, so:
\[ \frac{1}{2} \sin 2x \text{ varies between } -\frac{1}{2} \text{ and } \frac{1}{2} \]
Thus, the right-hand side \( 13 - \frac{1}{2} \) gives a minimum of \( 12.5 \) and a maximum of \( 13.5 \).

### Step 7: Conclusion
Since the maximum value of \( 4 \sin x - 3 \cos x \) is 5 and the minimum value of \( 13 - \frac{1}{2} \) is 12.5, we find that:
\[ 5 < 12.5 \]
This means that there is no \( x \) in the interval \( [0, 2\pi] \) for which the inequality holds true.

### Final Answer
Thus, the number of solutions of the inequality \( \sin x \cos x - 3 \cos x + 4 \sin x - 13 > 0 \) in the interval \( [0, 2\pi] \) is:
\[ \boxed{0} \]