The equation is $$\sin(3x) = \frac{\sqrt{2}}{2}$$. Let $$u = 3x$$. Since $$0 \le x < 2\pi$$, we have $$0 \le 3x < 6\pi$$, so $$0 \le u < 6\pi$$. The solutions for $$\sin(u) = \frac{\sqrt{2}}{2}$$ in $$[0, 2\pi)$$ are $$u = \frac{\pi}{4}$$ and $$u = \frac{3\pi}{4}$$. To find all solutions in $$[0, 6\pi)$$, we add multiples of $$2\pi$$. The solutions for $$u$$ are: $$\frac{\pi}{4}$$, $$\frac{3\pi}{4}$$, $$\frac{\pi}{4}+2\pi=\frac{9\pi}{4}$$, $$\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}$$, $$\frac{\pi}{4}+4\pi=\frac{17\pi}{4}$$, and $$\frac{3\pi}{4}+4\pi=\frac{19\pi}{4}$$. Each of these six values of $$u$$ gives a distinct value of $$x = u/3$$ in the interval $$[0, 2\pi)$$. Therefore, there are 6 solutions.

The equation simplifies to $$\csc^2(\theta) = 2$$. Taking the reciprocal of both sides gives $$\sin^2(\theta) = \frac{1}{2}$$. Taking the square root of both sides gives $$\sin(\theta) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$. This is true for all angles with a reference angle of $$\frac{\pi}{4}$$. The solutions in $$[0, 2\pi)$$ are $$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. These solutions are spaced $$\frac{\pi}{2}$$ apart. Thus, the general solution can be written compactly as $$\theta = \frac{\pi}{4} + \frac{k\pi}{2}$$.

First, isolate the cosine term: $$2\cos(2\theta - \frac{\pi}{2}) = -2$$, which simplifies to $$\cos(2\theta - \frac{\pi}{2}) = -1$$. The general solution for $$\cos(u) = -1$$ is $$u = \pi + 2k\pi$$. Let $$u = 2\theta - \frac{\pi}{2}$$. So, $$2\theta - \frac{\pi}{2} = \pi + 2k\pi$$. Add $$\frac{\pi}{2}$$ to both sides: $$2\theta = \frac{3\pi}{2} + 2k\pi$$. Finally, divide by 2: $$\theta = \frac{3\pi}{4} + k\pi$$.

The equation simplifies to $$\cos(\theta - \frac{\pi}{6}) = \frac{\sqrt{3}}{2}$$. Let $$u = \theta - \frac{\pi}{6}$$. As $$\theta$$ ranges from $$0$$ to $$2\pi$$, $$u$$ ranges from $$-\frac{\pi}{6}$$ to $$\frac{11\pi}{6}$$. The solutions for $$\cos(u) = \frac{\sqrt{3}}{2}$$ in this interval for $$u$$ are $$u = -\frac{\pi}{6}$$, $$u = \frac{\pi}{6}$$, and $$u = \frac{11\pi}{6}$$. Substituting back: 1) $$\theta - \frac{\pi}{6} = -\frac{\pi}{6} \implies \theta = 0$$. 2) $$\theta - \frac{\pi}{6} = \frac{\pi}{6} \implies \theta = \frac{\pi}{3}$$. 3) $$\theta - \frac{\pi}{6} = \frac{11\pi}{6} \implies \theta = 2\pi$$. All three solutions are in the specified interval $$[0, 2\pi]$$.

Set $$D(t) = 14.5$$ to solve for $$t$$: $$14.5 = 12 + 2.5\sin(\frac{2\pi}{365}(t-80))$$. Subtracting 12 gives $$2.5 = 2.5\sin(\frac{2\pi}{365}(t-80))$$, which simplifies to $$1 = \sin(\frac{2\pi}{365}(t-80))$$. The principal value for which sine is 1 is $$\frac{\pi}{2}$$. So, $$\frac{2\pi}{365}(t-80) = \frac{\pi}{2}$$. Dividing by $$2\pi$$ gives $$\frac{1}{365}(t-80) = \frac{1}{4}$$. Thus, $$t-80 = \frac{365}{4} = 91.25$$. Solving for $$t$$ gives $$t = 171.25$$. This corresponds to day 171.

Using the double-angle identity for cosine, $$\cos(2x) = 1 - 2\sin^2(x)$$, the equation becomes $$\sin(x) = 1 - 2\sin^2(x)$$. Rearranging gives the quadratic equation $$2\sin^2(x) + \sin(x) - 1 = 0$$. Factoring this equation yields $$(2\sin(x) - 1)(\sin(x) + 1) = 0$$. This implies $$\sin(x) = \frac{1}{2}$$ or $$\sin(x) = -1$$. On the interval $$[0, 2\pi)$$, $$\sin(x) = \frac{1}{2}$$ has two solutions, $$x=\frac{\pi}{6}$$ and $$x=\frac{5\pi}{6}$$. On the same interval, $$\sin(x) = -1$$ has one solution, $$x=\frac{3\pi}{2}$$. In total, there are 3 distinct solutions.

The inequality $$f(x) > 0$$ is equivalent to $$2\sin(x)-1 > 0$$, or $$\sin(x) > \frac{1}{2}$$. The solutions to $$\sin(x) = \frac{1}{2}$$ in $$[0, 2\pi]$$ are $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$. By examining the graph of $$y = \sin(x)$$ or the unit circle, we see that $$\sin(x)$$ is greater than $$\frac{1}{2}$$ for angles strictly between these two values. Therefore, the solution is the interval $$(\frac{\pi}{6}, \frac{5\pi}{6})$$.

The inequality simplifies to $$\sin^2(x) < \frac{3}{4}$$, which is equivalent to $$-\frac{\sqrt{3}}{2} < \sin(x) < \frac{\sqrt{3}}{2}$$. On the interval $$[0, \pi]$$, $$\sin(x)$$ is always non-negative, so the inequality becomes $$0 \le \sin(x) < \frac{\sqrt{3}}{2}$$. The solutions to $$\sin(x) = \frac{\sqrt{3}}{2}$$ on this interval are $$x = \frac{\pi}{3}$$ and $$x = \frac{2\pi}{3}$$. The inequality $$0 \le \sin(x) < \frac{\sqrt{3}}{2}$$ is satisfied when $$x$$ is in the interval $$[0, \frac{\pi}{3})$$ or in the interval $$(\frac{2\pi}{3}, \pi]$$.

This question tests AP Precalculus skills, specifically finding angles with negative sine values using reference angles. Given sin(θ) = -√2/2, we need angles where sine has this negative value. The reference angle is π/4 (since sin(π/4) = √2/2), and sine is negative in quadrants III and IV, giving θ = π + π/4 = 5π/4 and θ = 2π - π/4 = 7π/4. Choice B is correct because it identifies both solutions θ = 5π/4 and 7π/4 in [0, 2π). Choice A is incorrect because π/4 and 3π/4 have positive sine values (√2/2), not negative, showing a sign error. To help students: Use the unit circle to visualize where sine is negative (below the x-axis). Apply the reference angle systematically in quadrants III and IV for negative sine values.

This question tests AP Precalculus skills, specifically finding angles with given cosine values. The equation cos(θ) = 1/2 requires identifying angles in the restricted interval [0, π] where cosine equals one-half. On the unit circle, cos(θ) = 1/2 occurs at θ = π/3 (60°) and θ = 5π/3 (300°), but only π/3 lies within [0, π]. Choice A is correct because θ = π/3 is the only solution in the given interval. Choice B gives π/6 where cos(π/6) = √3/2, choice C gives 2π/3 where cos(2π/3) = -1/2, and choice D gives the degree measure instead of radians. To help students: memorize special angle values on the unit circle. Practice converting between degrees and radians, and always check that solutions fall within the specified interval.