91Ó°ÊÓ

First, we need to find the angles for which $$\sin z=\frac{\sqrt{2}}{2}$$, where $$z=3x$$. We know that $$\sin z = \frac{\sqrt{2}}{2}$$ for angles in the first and second quadrants: $$z = \frac{\pi}{4} + 2k\pi$$ and $$z = \frac{3\pi}{4} + 2k\pi$$, where k is an integer. Since we are looking for solutions for x, we need to replace z with 3x to find the general solution: $$3x = \frac{\pi}{4} + 2k\pi$$ and $$3x = \frac{3\pi}{4} + 2k\pi$$. Now, divide both sides of the equations by 3 to solve for x: $$x = \frac{\pi}{12} + \frac{2}{3}k\pi$$ and $$x = \frac{\pi}{4} + \frac{2}{3}k\pi$$.

Now, we need to find all specific solutions for x in the range $$0\leq x <2\pi$$. For the first equation $$x = \frac{\pi}{12} + \frac{2}{3}k\pi$$, substitute k = 0, 1, 2 and check if the corresponding x values fall within the given range: For k = 0: $$x = \frac{\pi}{12} + 0 = \frac{\pi}{12}$$ For k = 1: $$x = \frac{\pi}{12} + \frac{2}{3}\pi = \frac{5\pi}{12}$$ For k = 2: $$x = \frac{\pi}{12} + \frac{4}{3}\pi = \frac{9\pi}{12} > 2\pi$$ (outside the given range) For the second equation $$x = \frac{\pi}{4} + \frac{2}{3}k\pi$$, substitute k = 0, 1, 2 and check if the corresponding x values fall within the given range: For k = 0: $$x = \frac{\pi}{4} + 0 = \frac{\pi}{4}$$ For k = 1: $$x = \frac{\pi}{4} + \frac{2}{3}\pi = \frac{7\pi}{12}$$ For k = 2: $$x = \frac{\pi}{4} + \frac{4}{3}\pi = \frac{11\pi}{12}$$ Therefore, the specific solutions for x within the range $$0\leq x <2\pi$$ are $$x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{\pi}{4}, \frac{7\pi}{12}, \text{ and } \frac{11\pi}{12}$$.