If `4costheta-3sectheta=2tantheta`, then `theta` is equal to

Options

A. `npi + (-1)^(n)(pi/10)`
B. `npi + (-1)^(n)pi/6`
C. `npi -(-1)^(n) (3pi)/10`
D. `npi`

Answer

To solve the equation \( 4 \cos \theta - 3 \sec \theta = 2 \tan \theta \), we will follow these steps:

### Step 1: Rewrite the trigonometric functions in terms of sine and cosine.
We know that:
- \( \sec \theta = \frac{1}{\cos \theta} \)
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

Substituting these into the equation gives:
\[ 4 \cos \theta - 3 \cdot \frac{1}{\cos \theta} = 2 \cdot \frac{\sin \theta}{\cos \theta} \]

### Step 2: Multiply through by \( \cos \theta \) to eliminate the fractions.
This leads to:
\[ 4 \cos^2 \theta - 3 = 2 \sin \theta \]

### Step 3: Rearrange the equation.
Rearranging gives:
\[ 4 \cos^2 \theta - 2 \sin \theta - 3 = 0 \]

### Step 4: Use the Pythagorean identity.
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we substitute:
\[ 4(1 - \sin^2 \theta) - 2 \sin \theta - 3 = 0 \]
This simplifies to:
\[ 4 - 4 \sin^2 \theta - 2 \sin \theta - 3 = 0 \]
or
\[ -4 \sin^2 \theta - 2 \sin \theta + 1 = 0 \]

### Step 5: Multiply through by -1 to simplify.
This gives:
\[ 4 \sin^2 \theta + 2 \sin \theta - 1 = 0 \]

### Step 6: Solve the quadratic equation.
Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = 2, c = -1 \):
\[ \sin \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ = \frac{-2 \pm \sqrt{4 + 16}}{8} \]
\[ = \frac{-2 \pm \sqrt{20}}{8} \]
\[ = \frac{-2 \pm 2\sqrt{5}}{8} \]
\[ = \frac{-1 \pm \sqrt{5}}{4} \]

### Step 7: Find the values of \( \theta \).
Thus, we have:
1. \( \sin \theta = \frac{-1 + \sqrt{5}}{4} \)
2. \( \sin \theta = \frac{-1 - \sqrt{5}}{4} \) (not valid since sine values must be between -1 and 1)

### Step 8: Determine \( \theta \) from \( \sin \theta = \frac{-1 + \sqrt{5}}{4} \).
This value corresponds to angles:
- \( \theta = \sin^{-1} \left( \frac{-1 + \sqrt{5}}{4} \right) \)
- The general solutions can be expressed as:
\[ \theta = n\pi + (-1)^n \cdot \sin^{-1} \left( \frac{-1 + \sqrt{5}}{4} \right) \]

### Conclusion:
Thus, the final answer is:
\[ \theta = n\pi + (-1)^n \cdot \frac{\pi}{10} \quad \text{(Option 1)} \]