sin 9x + sin 8x + sin 7x = 0

Let's solve the equation $$sin 9x + sin 8x + sin 7x = 0.$$ We can group the first and third terms and use the sum-to-product formula: $$sin 9x + sin 7x = 2 sin \frac{9x + 7x}{2} cos \frac{9x - 7x}{2} = 2 sin 8x cos x.$$ Thus, the equation becomes $$2 sin 8x cos x + sin 8x = 0.$$ Factoring out $$\sin 8x$$, we get $$sin 8x (2 cos x + 1) = 0.$$ This equation is satisfied if either $$\sin 8x = 0$$ or $$2 cos x + 1 = 0$$. Case 1: $$\sin 8x = 0$$. This means $$8x = npi$$, where $$n$$ is an integer. Therefore, $$x = \frac{npi}{8},$$ where $$n$$ is an integer. Case 2: $$2 cos x + 1 = 0$$. This means $$\cos x = -\frac{1}{2}$$. The general solution for this is $$x = \pm \frac{2pi}{3} + 2kpi,$$ where $$k$$ is an integer. Therefore, the solutions are $$x = \frac{npi}{8}, \quad x = \frac{2pi}{3} + 2kpi, \quad x = -\frac{2pi}{3} + 2kpi,$$ where $$n$$ and $$k$$ are integers. Answer: $$x = \frac{npi}{8}, \quad x = \pm \frac{2pi}{3} + 2kpi$$