91Ó°ÊÓ

Since the equation is symmetric in sin and cos, we can factor sin(3x) and cos(3x). We have: $$\cos 3x(\sin^3 x) + \sin 3x (\cos^3 x) = 0$$ Notice that we can factor out a \(\sin 3x \cos 3x\) from the equation, and we get: $$\sin 3x \cos 3x(\sin^2 x + \cos^2 x) = 0$$ As we know that, $$\sin^2x + \cos^2x = 1$$, so the equation becomes, $$\sin 3x \cos 3x = 0$$

We can use the trigonometric identities for sin(3x) and cos(3x) as follows: $$\sin 3x = 3\sin x - 4\sin^3 x$$ $$\cos 3x = 4\cos^3 x - 3\cos x$$ Replace sin(3x) and cos(3x) in the equation: $$(3\sin x - 4\sin^3 x)(4\cos^3 x - 3\cos x) = 0$$

We now have two cases where the equation equals zero: Case 1: $$\sin 3x = 3\sin x - 4\sin^3 x = 0$$ Which can be expressed as: $$\sin x (3 - 4\sin^2 x) = 0$$ So, for this equation to be zero, either: $$\sin x = 0 \Rightarrow x = k\pi$$ or $$4\sin^2 x - 3 = 0 \Rightarrow \sin^2 x = \frac{3}{4} \Rightarrow \sin x = \pm\frac{\sqrt{3}}{2}$$ Which leads to $$x = \begin{cases} \frac{\pi}{3} + 2k\pi \\ \frac{2\pi}{3} + 2k\pi \\ \frac{4\pi}{3} + 2k\pi \\ \frac{5\pi}{3} + 2k\pi \end{cases}$$ Case 2: $$\cos 3x = 4\cos^3 x - 3\cos x = 0$$ Which can be expressed as: $$\cos x (4\cos^2 x - 3) = 0$$ So, for this equation to be zero, either: $$\cos x = 0 \Rightarrow x = \frac{\pi}{2}+k\pi$$ or $$4\cos^2 x - 3 = 0 \Rightarrow \cos^2 x = \frac{3}{4} \Rightarrow \cos x = \pm\frac{\sqrt{3}}{2}$$ Which leads to $$x = \begin{cases} \frac{\pi}{6} + 2k\pi \\ \frac{7\pi}{6} + 2k\pi \\ \frac{11\pi}{6} + 2k\pi \\ \frac{5\pi}{2} + 2k\pi \end{cases}$$ Combining the solutions from both cases, we obtain the general solution for x: $$x = \begin{cases} k\pi \\ \frac{\pi}{2}+k\pi \\ \frac{\pi}{6} + 2k\pi \\ \frac{7\pi}{6} + 2k\pi \\ \frac{11\pi}{6} + 2k\pi \\ \frac{\pi}{3} + 2k\pi \\ \frac{2\pi}{3} + 2k\pi \\ \frac{4\pi}{3} + 2k\pi \\ \frac{5\pi}{3} + 2k\pi \end{cases} , k \in \mathbb{Z}$$