Question: cos5x+sinx+sin7x-cos3x=0
1. **State the problem:** Solve the trigonometric equation $$\cos 5x + \sin x + \sin 7x - \cos 3x = 0$$.
2. **Use trigonometric identities:** Recall the sum-to-product formulas:
$$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$
$$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$
3. **Apply sum-to-product to the sine terms:**
$$\sin x + \sin 7x = 2 \sin \left( \frac{x + 7x}{2} \right) \cos \left( \frac{x - 7x}{2} \right) = 2 \sin 4x \cos (-3x)$$
Since $$\cos (-3x) = \cos 3x$$, this becomes:
$$2 \sin 4x \cos 3x$$
4. **Apply sum-to-product to the cosine terms:**
$$\cos 5x - \cos 3x = -2 \sin \left( \frac{5x + 3x}{2} \right) \sin \left( \frac{5x - 3x}{2} \right) = -2 \sin 4x \sin x$$
5. **Rewrite the original equation:**
$$\cos 5x + \sin x + \sin 7x - \cos 3x = (\cos 5x - \cos 3x) + (\sin x + \sin 7x) = -2 \sin 4x \sin x + 2 \sin 4x \cos 3x = 0$$
6. **Factor out common terms:**
$$2 \sin 4x (\cos 3x - \sin x) = 0$$
7. **Set each factor equal to zero:**
- $$\sin 4x = 0$$
- $$\cos 3x - \sin x = 0$$
8. **Solve $$\sin 4x = 0$$:**
$$4x = n\pi, \quad n \in \mathbb{Z}$$
$$x = \frac{n\pi}{4}$$
9. **Solve $$\cos 3x = \sin x$$:**
Rewrite $$\sin x$$ as $$\cos \left( \frac{\pi}{2} - x \right)$$:
$$\cos 3x = \cos \left( \frac{\pi}{2} - x \right)$$
10. **Use cosine equality:**
$$\cos A = \cos B \implies A = 2k\pi \pm B, \quad k \in \mathbb{Z}$$
So,
$$3x = 2k\pi \pm \left( \frac{\pi}{2} - x \right)$$
11. **Solve for $$x$$:**
- Case 1: $$3x = 2k\pi + \frac{\pi}{2} - x \implies 4x = 2k\pi + \frac{\pi}{2} \implies x = \frac{2k\pi + \frac{\pi}{2}}{4} = \frac{k\pi}{2} + \frac{\pi}{8}$$
- Case 2: $$3x = 2k\pi - \frac{\pi}{2} + x \implies 2x = 2k\pi - \frac{\pi}{2} \implies x = k\pi - \frac{\pi}{4}$$
12. **Final solution set:**
$$x = \frac{n\pi}{4}, \quad x = \frac{k\pi}{2} + \frac{\pi}{8}, \quad x = k\pi - \frac{\pi}{4}, \quad n,k \in \mathbb{Z}$$