If `|(cos theta,-1,1),(cos2 theta,4,3),(2,7,7)|=0`, then the number of values of `theta` in `[0, 1pi]` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem given by the determinant condition \(|(cos \theta, -1, 1), (cos 2\theta, 4, 3), (2, 7, 7)| = 0\), we will follow these steps:

### Step 1: Calculate the Determinant
The determinant of a 3x3 matrix can be calculated using the formula:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \begin{vmatrix} \cos \theta & -1 & 1 \\ \cos 2\theta & 4 & 3 \\ 2 & 7 & 7 \end{vmatrix} \]

We can expand this determinant:
\[ D = \cos \theta \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} - (-1) \begin{vmatrix} \cos 2\theta & 3 \\ 2 & 7 \end{vmatrix} + 1 \begin{vmatrix} \cos 2\theta & 4 \\ 2 & 7 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} = (4 \cdot 7) - (3 \cdot 7) = 28 - 21 = 7\)
2. \(\begin{vmatrix} \cos 2\theta & 3 \\ 2 & 7 \end{vmatrix} = (\cos 2\theta \cdot 7) - (3 \cdot 2) = 7\cos 2\theta - 6\)
3. \(\begin{vmatrix} \cos 2\theta & 4 \\ 2 & 7 \end{vmatrix} = (\cos 2\theta \cdot 7) - (4 \cdot 2) = 7\cos 2\theta - 8\)

### Step 3: Substitute Back into the Determinant
Substituting these back into the determinant expression:
\[ D = \cos \theta \cdot 7 + (7\cos 2\theta - 6) + (7\cos 2\theta - 8) \]
Combine like terms:
\[ D = 7\cos \theta + 14\cos 2\theta - 14 \]

### Step 4: Set the Determinant to Zero
Setting the determinant equal to zero:
\[ 7\cos \theta + 14\cos 2\theta - 14 = 0 \]
Dividing through by 7 gives:
\[ \cos \theta + 2\cos 2\theta - 2 = 0 \]

### Step 5: Use the Cosine Double Angle Identity
Using the identity \(\cos 2\theta = 2\cos^2 \theta - 1\):
\[ \cos \theta + 2(2\cos^2 \theta - 1) - 2 = 0 \]
This simplifies to:
\[ \cos \theta + 4\cos^2 \theta - 2 - 2 = 0 \]
Thus:
\[ 4\cos^2 \theta + \cos \theta - 4 = 0 \]

### Step 6: Solve the Quadratic Equation
Let \(x = \cos \theta\):
\[ 4x^2 + x - 4 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} = \frac{-1 \pm \sqrt{1 + 64}}{8} = \frac{-1 \pm \sqrt{65}}{8} \]

### Step 7: Find the Values of \(\theta\)
The solutions for \(x = \cos \theta\) are:
1. \(x_1 = \frac{-1 + \sqrt{65}}{8}\)
2. \(x_2 = \frac{-1 - \sqrt{65}}{8}\)

### Step 8: Determine Validity of Solutions
1. For \(x_1\), check if \(-1 \leq x_1 \leq 1\):
- Calculate \(\sqrt{65} \approx 8.06\), so \(x_1 \approx \frac{7.06}{8} \approx 0.8825\) (valid).
2. For \(x_2\), since \(\sqrt{65} > 1\), \(x_2\) is negative and less than -1 (invalid).

### Step 9: Count the Values of \(\theta\)
Since \(x_1\) is valid, we find the angles:
\[ \theta = \cos^{-1}(x_1) \quad \text{and} \quad \theta = 2\pi - \cos^{-1}(x_1) \]
Both angles are within the interval \([0, \pi]\).

### Final Answer
Thus, the number of values of \(\theta\) in the interval \([0, \pi]\) is **2**.

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