If `5(tan^2x - cos^2x)=2cos 2x + 9`, then the value of cos4x is

Options

A. `- (7)/(9)`
B. `- (3)/(5)`
C. `(1)/(3)`
D. `(2)/(9)`

Explanation

To solve the equation \( 5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9 \) and find the value of \( \cos 4x \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \tan^2 x = \sec^2 x - 1 \). Therefore, we can rewrite \( \tan^2 x \) in terms of \( \cos^2 x \):
\[ \tan^2 x = \sec^2 x - 1 = \frac{1}{\cos^2 x} - 1 \]
Substituting this into the equation gives:
\[ 5\left(\frac{1}{\cos^2 x} - 1 - \cos^2 x\right) = 2\cos 2x + 9 \]

### Step 2: Simplify the left side
The left side simplifies to:
\[ 5\left(\frac{1 - \cos^2 x - \cos^4 x}{\cos^2 x}\right) = 5\left(\frac{1 - \cos^2 x - \cos^4 x}{\cos^2 x}\right) \]
This can be rewritten as:
\[ \frac{5(1 - \cos^2 x - \cos^4 x)}{\cos^2 x} \]

### Step 3: Rewrite \( \cos 2x \)
Using the identity \( \cos 2x = 2\cos^2 x - 1 \), we can substitute into the equation:
\[ 5\left(\frac{1 - \cos^2 x - \cos^4 x}{\cos^2 x}\right) = 2(2\cos^2 x - 1) + 9 \]
This simplifies to:
\[ 5\left(\frac{1 - \cos^2 x - \cos^4 x}{\cos^2 x}\right) = 4\cos^2 x - 2 + 9 \]
\[ 5\left(\frac{1 - \cos^2 x - \cos^4 x}{\cos^2 x}\right) = 4\cos^2 x + 7 \]

### Step 4: Substitute \( t = \cos^2 x \)
Let \( t = \cos^2 x \). The equation becomes:
\[ 5\left(\frac{1 - t - t^2}{t}\right) = 4t + 7 \]
Cross-multiplying gives:
\[ 5(1 - t - t^2) = (4t + 7)t \]
Expanding both sides results in:
\[ 5 - 5t - 5t^2 = 4t^2 + 7t \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ 5 - 5t - 5t^2 - 4t^2 - 7t = 0 \]
This simplifies to:
\[ -9t^2 - 12t + 5 = 0 \]
Multiplying through by -1:
\[ 9t^2 + 12t - 5 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 9 \cdot (-5)}}{2 \cdot 9} \]
Calculating the discriminant:
\[ t = \frac{-12 \pm \sqrt{144 + 180}}{18} = \frac{-12 \pm \sqrt{324}}{18} = \frac{-12 \pm 18}{18} \]
This gives us two solutions:
\[ t = \frac{6}{18} = \frac{1}{3} \quad \text{and} \quad t = \frac{-30}{18} \quad \text{(invalid since } t = \cos^2 x \text{ must be non-negative)} \]

### Step 7: Find \( \cos 2x \) and \( \cos 4x \)
Since \( t = \cos^2 x = \frac{1}{3} \):
\[ \cos 2x = 2\cos^2 x - 1 = 2 \cdot \frac{1}{3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3} \]
Now, using \( \cos 4x = 2\cos^2 2x - 1 \):
\[ \cos 4x = 2\left(-\frac{1}{3}\right)^2 - 1 = 2 \cdot \frac{1}{9} - 1 = \frac{2}{9} - 1 = \frac{2}{9} - \frac{9}{9} = -\frac{7}{9} \]

### Final Answer
Thus, the value of \( \cos 4x \) is:
\[ \boxed{-\frac{7}{9}} \]