1. The problem is to solve a very challenging trigonometry question that is considered extremely difficult. 2. One example of a tough trigonometry problem is to solve the equation $$\sin(x) + \sin(2x) + \sin(3x) = 0$$ for all values of $x$. 3. The formula used involves sum-to-product identities and solving transcendental equations. 4. Using sum-to-product identities: $$\sin(x) + \sin(3x) = 2 \sin\left(\frac{x+3x}{2}\right) \cos\left(\frac{x-3x}{2}\right) = 2 \sin(2x) \cos(-x) = 2 \sin(2x) \cos(x)$$ 5. Substitute back: $$2 \sin(2x) \cos(x) + \sin(2x) = \sin(2x)(2 \cos(x) + 1) = 0$$ 6. Set each factor to zero: - $\sin(2x) = 0$ gives $2x = n\pi \Rightarrow x = \frac{n\pi}{2}$ - $2 \cos(x) + 1 = 0$ gives $\cos(x) = -\frac{1}{2}$ which implies $x = \pm \frac{2\pi}{3} + 2n\pi$ 7. Therefore, the solutions are: $$x = \frac{n\pi}{2} \quad \text{or} \quad x = \pm \frac{2\pi}{3} + 2n\pi, \quad n \in \mathbb{Z}$$ This problem requires knowledge of trigonometric identities and solving transcendental equations, making it quite challenging.