For `0ltxlt(pi)/(6),` all the values of `tan^(2)3x cos^(2)x-4tan3xsin2x+16sin^(2)x` lie in the interval

Options

A. `(0,(121)/(36))`
B. `(1,(121)/(9))`
C. `(-1,0)`
D. none of these

Explanation

To solve the expression \( \tan^2(3x) \cos^2(x) - 4 \tan(3x) \sin(2x) + 16 \sin^2(x) \) for \( 0 < x < \frac{\pi}{6} \) and determine the interval of all its values, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = \tan^2(3x) \cos^2(x) - 4 \tan(3x) \sin(2x) + 16 \sin^2(x) \]
We know that \( \sin(2x) = 2 \sin(x) \cos(x) \), so we can substitute this into the expression:
\[ y = \tan^2(3x) \cos^2(x) - 8 \tan(3x) \sin(x) \cos(x) + 16 \sin^2(x) \]

### Step 2: Factor out common terms
Next, we can factor out \( \sin^2(x) \):
\[ y = \sin^2(x) \left( \frac{\tan^2(3x)}{\cos^2(x)} - 8 \frac{\tan(3x)}{\sin(x) \cos(x)} + 16 \right) \]
This simplifies to:
\[ y = \sin^2(x) \left( \tan^2(3x) - 8 \tan(3x) \frac{1}{\sin(x)} + 16 \cos^2(x) \right) \]

### Step 3: Substitute \( \tan(3x) \)
Let \( t = \tan(3x) \). Then, we can rewrite the expression as:
\[ y = \sin^2(x) \left( t^2 - 8t + 16 \cos^2(x) \right) \]

### Step 4: Analyze the quadratic
The quadratic \( t^2 - 8t + 16 \cos^2(x) \) can be analyzed using the quadratic formula:
\[ t = \frac{8 \pm \sqrt{64 - 64 \cos^2(x)}}{2} = 4 \pm 4 \sqrt{1 - \cos^2(x)} = 4 \pm 4 \sin(x) \]
Thus, the roots of the quadratic are:
\[ t_1 = 4 + 4 \sin(x), \quad t_2 = 4 - 4 \sin(x) \]

### Step 5: Determine the range of \( t \)
For \( x \) in the interval \( (0, \frac{\pi}{6}) \):
- \( \sin(x) \) ranges from \( 0 \) to \( \frac{1}{2} \).
- Therefore, \( t_1 \) ranges from \( 4 \) to \( 6 \) and \( t_2 \) ranges from \( 4 \) to \( 2 \).

### Step 6: Find the minimum and maximum values of \( y \)
The minimum value of \( y \) occurs when \( \sin^2(x) = 0 \) (which gives \( y = 0 \)) and the maximum occurs when \( t = 6 \) (which gives the maximum value of \( y \)):
\[ y_{\text{max}} = \sin^2(x) \cdot (6^2) = \sin^2(x) \cdot 36 \]
At \( x = \frac{\pi}{6} \):
\[ \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus:
\[ y_{\text{max}} = 36 \cdot \frac{1}{4} = 9 \]

### Conclusion
The values of \( y \) lie in the interval:
\[ [0, 9] \]