`5(tan^(2)x-cos^(2)x) = 2cos2x +9` , then the value of cos4x is:

Options

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

Answer

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:
\[ \tan^2 x = \sec^2 x - 1 \]
Substituting this into the equation gives:
\[ 5((\sec^2 x - 1) - \cos^2 x) = 2 \cos 2x + 9 \]

### Step 2: Substitute \( \sec^2 x \)
Since \( \sec^2 x = \frac{1}{\cos^2 x} \), we can rewrite the equation as:
\[ 5\left(\frac{1}{\cos^2 x} - 1 - \cos^2 x\right) = 2 \cos 2x + 9 \]

### Step 3: Use the identity for \( \cos 2x \)
The identity for \( \cos 2x \) is:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Substituting this into the equation gives:
\[ 5\left(\frac{1}{\cos^2 x} - 1 - \cos^2 x\right) = 2(2 \cos^2 x - 1) + 9 \]

### Step 4: Simplify the right-hand side
The right-hand side simplifies to:
\[ 4 \cos^2 x - 2 + 9 = 4 \cos^2 x + 7 \]

### Step 5: Set the equation
Now we have:
\[ 5\left(\frac{1}{\cos^2 x} - 1 - \cos^2 x\right) = 4 \cos^2 x + 7 \]

### Step 6: Multiply through by \( \cos^2 x \) to eliminate the fraction
\[ 5(1 - \cos^2 x - \cos^4 x) = (4 \cos^2 x + 7) \cos^2 x \]
Expanding both sides gives:
\[ 5 - 5 \cos^2 x - 5 \cos^4 x = 4 \cos^4 x + 7 \cos^2 x \]

### Step 7: Rearrange the equation
Rearranging terms results in:
\[ -5 \cos^4 x - 4 \cos^4 x - 5 \cos^2 x - 7 \cos^2 x + 5 = 0 \]
Combining like terms:
\[ -9 \cos^4 x - 12 \cos^2 x + 5 = 0 \]

### Step 8: Substitute \( t = \cos^2 x \)
Let \( t = \cos^2 x \):
\[ -9t^2 - 12t + 5 = 0 \]
Multiplying through by -1 gives:
\[ 9t^2 + 12t - 5 = 0 \]

### Step 9: Use the quadratic formula
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:
\[ t = \frac{-12 \pm \sqrt{144 + 180}}{18} = \frac{-12 \pm \sqrt{324}}{18} = \frac{-12 \pm 18}{18} \]

### Step 10: Solve for \( t \)
Calculating the two possible values for \( t \):
1. \( t = \frac{6}{18} = \frac{1}{3} \)
2. \( t = \frac{-30}{18} \) (not valid since \( t \) must be non-negative)

Thus, \( \cos^2 x = \frac{1}{3} \).

### Step 11: Find \( \cos 2x \)
Using the identity \( \cos 2x = 2 \cos^2 x - 1 \):
\[ \cos 2x = 2 \cdot \frac{1}{3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3} \]

### Step 12: Find \( \cos 4x \)
Using the identity \( \cos 4x = 2 \cos^2 2x - 1 \):
First, calculate \( \cos^2 2x \):
\[ \cos^2 2x = \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \]
Now substitute into the identity:
\[ \cos 4x = 2 \cdot \frac{1}{9} - 1 = \frac{2}{9} - 1 = -\frac{7}{9} \]

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