The sum of all possible values of $$\theta \in [-\pi, 2\pi]$$, for which $$\frac{1 + i\cos\theta}{1 - 2i\cos\theta}$$ is purely imaginary, is equal to :

We need to find the sum of all values of $$\theta \in [-\pi, 2\pi]$$ for which $$\frac{1 + i\cos\theta}{1 - 2i\cos\theta}$$ is purely imaginary.

For a complex number to be purely imaginary, its real part must be zero.

Let us rationalize: multiply numerator and denominator by the conjugate of the denominator.

$$\frac{1 + i\cos\theta}{1 - 2i\cos\theta} \times \frac{1 + 2i\cos\theta}{1 + 2i\cos\theta} = \frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{1 + 4\cos^2\theta}$$

Expanding the numerator:

$$(1)(1) + (1)(2i\cos\theta) + (i\cos\theta)(1) + (i\cos\theta)(2i\cos\theta)$$

$$= 1 + 2i\cos\theta + i\cos\theta + 2i^2\cos^2\theta$$

$$= 1 + 3i\cos\theta - 2\cos^2\theta$$

$$= (1 - 2\cos^2\theta) + 3i\cos\theta$$

For the expression to be purely imaginary, the real part must be zero:

$$1 - 2\cos^2\theta = 0$$

$$\cos^2\theta = \frac{1}{2}$$

$$\cos\theta = \pm\frac{1}{\sqrt{2}}$$

We also need to verify the imaginary part is non-zero, i.e., $$\cos\theta \neq 0$$, which is satisfied.

The values of $$\theta$$ in $$[-\pi, 2\pi]$$:

$$\cos\theta = \frac{1}{\sqrt{2}}$$: $$\theta = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{7\pi}{4}$$

$$\cos\theta = -\frac{1}{\sqrt{2}}$$: $$\theta = \frac{3\pi}{4}, -\frac{3\pi}{4}, \frac{5\pi}{4}$$

Sum = $$-\frac{\pi}{4} + \frac{\pi}{4} + \frac{7\pi}{4} + \frac{3\pi}{4} + (-\frac{3\pi}{4}) + \frac{5\pi}{4}$$

$$= 0 + \frac{7\pi}{4} + \frac{3\pi}{4} - \frac{3\pi}{4} + \frac{5\pi}{4} = \frac{7\pi + 5\pi}{4} = \frac{12\pi}{4} = 3\pi$$

The correct answer is Option 1: $$3\pi$$.