If $$2\cos\theta + \sin\theta = 1$$ ($$\theta \neq \frac{\pi}{2}$$), then $$7\cos\theta + 6\sin\theta$$ is equal to:

We are given that $$2\cos\theta + \sin\theta = 1$$ where $$\theta \neq \frac{\pi}{2}$$. We need to find $$7\cos\theta + 6\sin\theta$$.

Step 1: Use the half-angle substitution.

Let $$t = \tan\frac{\theta}{2}$$. Then using the standard identities:

$$\cos\theta = \frac{1 - t^2}{1 + t^2}$$, $$\quad \sin\theta = \frac{2t}{1 + t^2}$$

Step 2: Substitute into the given equation.

$$2 \cdot \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2} = 1$$

Multiplying both sides by $$(1 + t^2)$$:

$$2(1 - t^2) + 2t = 1 + t^2$$

$$2 - 2t^2 + 2t = 1 + t^2$$

$$3t^2 - 2t - 1 = 0$$

Step 3: Solve for $$t$$.

Using the quadratic formula:

$$t = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6}$$

$$t = 1 \quad \text{or} \quad t = -\frac{1}{3}$$

If $$t = \tan\frac{\theta}{2} = 1$$, then $$\frac{\theta}{2} = \frac{\pi}{4}$$, so $$\theta = \frac{\pi}{2}$$. But $$\theta \neq \frac{\pi}{2}$$ is given, so we reject this.

Therefore $$t = -\frac{1}{3}$$.

Step 4: Find $$\cos\theta$$ and $$\sin\theta$$.

$$\cos\theta = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} = \frac{\frac{8}{9}}{\frac{10}{9}} = \frac{4}{5}$$

$$\sin\theta = \frac{2 \cdot (-\frac{1}{3})}{1 + \frac{1}{9}} = \frac{-\frac{2}{3}}{\frac{10}{9}} = -\frac{3}{5}$$

Step 5: Verify. $$2\cos\theta + \sin\theta = 2 \cdot \frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{8}{5} - \frac{3}{5} = \frac{5}{5} = 1$$ ✓

Step 6: Compute the required expression.

$$7\cos\theta + 6\sin\theta = 7 \cdot \frac{4}{5} + 6 \cdot \left(-\frac{3}{5}\right) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$$

The correct answer is Option B: $$2$$.