If `32"tan"^(8)theta" = 2"cos"^(2) alpha- 3"cos" alpha " and "3"cos" 2 theta = 1, "then" alpha =`

Options

A. `2n pi, n in Z`
B. `2n pi +- (2pi)/(3), n in Z`
C. `2n pi +- (pi)/(3), n in Z`
D. `n pi +- (pi)/(3), n in Z`

Answer

To solve the problem, we need to follow these steps systematically:

### Step 1: Solve for \( \cos 2\theta \)

Given the equation:
\[ 3 \cos 2\theta = 1 \]
We can isolate \( \cos 2\theta \):
\[ \cos 2\theta = \frac{1}{3} \]

**Hint:** Use the identity for \( \cos 2\theta \) in terms of \( \tan \theta \).

### Step 2: Use the double angle formula

Using the double angle formula:
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
Substituting \( \cos 2\theta = \frac{1}{3} \):
\[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1}{3} \]

**Hint:** Cross-multiply to eliminate the fraction.

### Step 3: Cross-multiply

Cross-multiplying gives:
\[ 3(1 - \tan^2 \theta) = 1 + \tan^2 \theta \]
Expanding this, we get:
\[ 3 - 3\tan^2 \theta = 1 + \tan^2 \theta \]

**Hint:** Rearrange the equation to isolate terms involving \( \tan^2 \theta \).

### Step 4: Rearranging the equation

Rearranging the equation leads to:
\[ 3 - 1 = 3\tan^2 \theta + \tan^2 \theta \]
This simplifies to:
\[ 2 = 4\tan^2 \theta \]

**Hint:** Solve for \( \tan^2 \theta \).

### Step 5: Solve for \( \tan^2 \theta \)

Dividing both sides by 4 gives:
\[ \tan^2 \theta = \frac{1}{2} \]

**Hint:** Substitute this value into the next equation.

### Step 6: Substitute into the original equation

Now, substitute \( \tan^2 \theta \) into the original equation:
\[ 32 \tan^8 \theta = 2 \cos^2 \alpha - 3 \cos \alpha \]
We know \( \tan^8 \theta = \left(\tan^2 \theta\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \):
\[ 32 \cdot \frac{1}{16} = 2 \cos^2 \alpha - 3 \cos \alpha \]
This simplifies to:
\[ 2 = 2 \cos^2 \alpha - 3 \cos \alpha \]

**Hint:** Rearrange this into a standard quadratic form.

### Step 7: Rearranging into a quadratic equation

Rearranging gives:
\[ 2 \cos^2 \alpha - 3 \cos \alpha - 2 = 0 \]

**Hint:** Use the quadratic formula to solve for \( \cos \alpha \).

### Step 8: Apply the quadratic formula

Using the quadratic formula \( \cos \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = -2 \):
\[ \cos \alpha = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \cos \alpha = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4} \]

**Hint:** Calculate the two possible values for \( \cos \alpha \).

### Step 9: Calculate the values

Calculating gives:
1. \( \cos \alpha = \frac{8}{4} = 2 \) (not possible since \( \cos \) must be between -1 and 1)
2. \( \cos \alpha = \frac{-2}{4} = -\frac{1}{2} \)

**Hint:** Determine the angle corresponding to \( \cos \alpha = -\frac{1}{2} \).

### Step 10: Find \( \alpha \)

The value \( \cos \alpha = -\frac{1}{2} \) corresponds to:
\[ \alpha = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \alpha = \frac{4\pi}{3} + 2n\pi \]
where \( n \) is any integer.

**Final Answer:**
\[ \alpha = 2n\pi \pm \frac{2\pi}{3} \]