A line in the 3-dimensional space makes an angle $$\theta$$ ($$0 < \theta \leq \frac{\pi}{2}$$) with both the X and Y-axes. Then, the set of all values of $$\theta$$ is in the interval:

Consider a line in 3-dimensional space with direction cosines $$l$$, $$m$$, and $$n$$, which represent the cosines of the angles the line makes with the X-axis, Y-axis, and Z-axis, respectively. Given that the line makes an angle $$\theta$$ with both the X-axis and Y-axis, we have $$l = \cos \theta$$ and $$m = \cos \theta$$.

The fundamental property of direction cosines states that $$l^2 + m^2 + n^2 = 1$$. Substituting the values, we get:

$$(\cos \theta)^2 + (\cos \theta)^2 + n^2 = 1$$

Simplifying, this becomes:

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

Rearranging for $$n^2$$:

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

Using the double-angle identity $$\cos(2\theta) = 2 \cos^2 \theta - 1$$, we can rewrite $$1 - 2 \cos^2 \theta$$ as:

$$1 - 2 \cos^2 \theta = -\left(2 \cos^2 \theta - 1\right) = -\cos(2\theta)$$

Thus,

$$n^2 = -\cos(2\theta)$$

Since $$n^2$$ represents the square of a real number, it must be non-negative:

$$n^2 \geq 0 \quad \Rightarrow \quad -\cos(2\theta) \geq 0 \quad \Rightarrow \quad \cos(2\theta) \leq 0$$

The condition $$\cos(2\theta) \leq 0$$ holds when $$2\theta$$ lies in the interval where cosine is non-positive. Given that $$\theta$$ is in $$(0, \pi/2]$$, the range for $$2\theta$$ is $$(0, \pi]$$. Cosine is non-positive in the interval $$[\pi/2, \pi]$$, so:

$$\frac{\pi}{2} \leq 2\theta \leq \pi$$

Dividing by 2:

$$\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$$

Now, verify the endpoints:

• At $$\theta = \pi/4$$, $$l = \cos(\pi/4) = \sqrt{2}/2$$, $$m = \sqrt{2}/2$$. Then $$n^2 = 1 - 2 \left(\sqrt{2}/2\right)^2 = 1 - 2 \times (1/2) = 1 - 1 = 0$$, so $$n = 0$$, which is valid.
• At $$\theta = \pi/2$$, $$l = \cos(\pi/2) = 0$$, $$m = 0$$. Then $$n^2 = 1 - 2 \times 0^2 = 1$$, so $$n = \pm 1$$, which is valid.

Values outside this interval are invalid. For example, if $$\theta = \pi/6$$ (less than $$\pi/4$$), then $$l = \cos(\pi/6) = \sqrt{3}/2$$, $$m = \sqrt{3}/2$$, and $$n^2 = 1 - 2 \left(\sqrt{3}/2\right)^2 = 1 - 2 \times (3/4) = 1 - 3/2 = -1/2 < 0$$, which is impossible.

Thus, the set of all possible values of $$\theta$$ is the closed interval $$[\pi/4, \pi/2]$$. Comparing with the options:

• A. $$\left(\frac{\pi}{3}, \frac{\pi}{2}\right]$$ excludes $$\pi/3$$ and $$\pi/4$$, but $$\pi/4$$ is valid.
• B. $$\left(0, \frac{\pi}{4}\right]$$ includes values less than $$\pi/4$$ (e.g., $$\pi/6$$) which are invalid.
• C. $$\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$$ matches the derived interval.
• D. $$\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$$ includes $$\pi/6$$ which is invalid.

Hence, the correct answer is Option C.