Solve the trigonometric equation:

Okay, let's solve the trigonometric equation: Given equation: $$2\sin{\left(-\frac{5x}{12}\right)} = -\sqrt{2}$$ First, divide both sides by 2: $$\sin{\left(-\frac{5x}{12}\right)} = -\frac{\sqrt{2}}{2}$$ Since $$\sin(-\alpha) = -\sin(\alpha)$$, we can rewrite the equation as: $$-\sin{\left(\frac{5x}{12}\right)} = -\frac{\sqrt{2}}{2}$$ Multiply both sides by -1: $$\sin{\left(\frac{5x}{12}\right)} = \frac{\sqrt{2}}{2}$$ The general solution for $$\sin(\theta) = \frac{\sqrt{2}}{2}$$ is: $$\theta = \frac{\pi}{4} + 2\pi k$$ or $$\theta = \frac{3\pi}{4} + 2\pi k$$ where $$k$$ is an integer. Therefore, we have two cases: Case 1: $$\frac{5x}{12} = \frac{\pi}{4} + 2\pi k$$ Multiply both sides by $$\frac{12}{5}$$: $$x = \frac{12}{5} \left(\frac{\pi}{4} + 2\pi k\right) = \frac{3\pi}{5} + \frac{24\pi k}{5}$$ Case 2: $$\frac{5x}{12} = \frac{3\pi}{4} + 2\pi k$$ Multiply both sides by $$\frac{12}{5}$$: $$x = \frac{12}{5} \left(\frac{3\pi}{4} + 2\pi k\right) = \frac{9\pi}{5} + \frac{24\pi k}{5}$$ Thus, the general solutions are: $$x = \frac{3\pi}{5} + \frac{24\pi k}{5}$$ $$x = \frac{9\pi}{5} + \frac{24\pi k}{5}$$ where $$k$$ is an integer. Answer: The solutions are $$x = \frac{3\pi}{5} + \frac{24\pi k}{5}$$ and $$x = \frac{9\pi}{5} + \frac{24\pi k}{5}$$, where *k* is an integer.