$$ (5.5) \sin 3 x=\cos x $$ (5.6) $$ \sin x=\sin 43^{\circ}+\sin 17^{\circ} $$

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**(5.5) Solve $$ \sin 3x = \cos x $$:**

1. Express $$\cos x$$ in terms of sine:
   $$
   \cos x = \sin\left(\frac{\pi}{2} - x\right).
   $$
   Then the equation becomes:
   $$
   \sin 3x = \sin\left(\frac{\pi}{2} - x\right).
   $$

2. Use the general solution for $$\sin A = \sin B$$, which is:
   $$
   A = B + 2k\pi \quad \text{or} \quad A = \pi - B + 2k\pi,\quad k\in\mathbb{Z}.
   $$

3. **Case 1:** 
   
   $$
   3x = \frac{\pi}{2} - x + 2k\pi.
   $$
   Solve for $$x$$:
   $$
   3x + x = \frac{\pi}{2} + 2k\pi \quad\Longrightarrow\quad 4x = \frac{\pi}{2} + 2k\pi,
   $$
   $$
   x = \frac{\pi}{8} + \frac{k\pi}{2}.
   $$

4. **Case 2:** 
   
   $$
   3x = \pi - \left(\frac{\pi}{2} - x\right) + 2k\pi.
   $$
   Simplify inside the parentheses:
   $$
   \pi - \left(\frac{\pi}{2} - x\right) = \frac{\pi}{2} + x.
   $$
   So the equation becomes:
   $$
   3x = \frac{\pi}{2} + x + 2k\pi,
   $$
   $$
   3x - x = \frac{\pi}{2} + 2k\pi \quad\Longrightarrow\quad 2x = \frac{\pi}{2} + 2k\pi,
   $$
   $$
   x = \frac{\pi}{4} + k\pi.
   $$

**Final solutions for (5.5):**
$$
x = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{\pi}{4} + k\pi,\quad k\in\mathbb{Z}.
$$

---

**(5.6) Solve $$ \sin x = \sin 43^\circ + \sin 17^\circ $$:**

1. Use the sum-to-product formula:
   $$
   \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right).
   $$
   Let $$ A = 43^\circ $$ and $$ B = 17^\circ $$. Then:
   $$
   \frac{A+B}{2} = \frac{43^\circ + 17^\circ}{2} = 30^\circ,
   $$
   $$
   \frac{A-B}{2} = \frac{43^\circ - 17^\circ}{2} = 13^\circ.
   $$
   Thus:
   $$
   \sin 43^\circ + \sin 17^\circ = 2\sin 30^\circ \cos 13^\circ.
   $$
2. Since $$\sin 30^\circ = \frac{1}{2}$$, we have:
   $$
   2\sin 30^\circ \cos 13^\circ = 2\cdot\frac{1}{2}\cos 13^\circ = \cos 13^\circ.
   $$
   Hence the equation becomes:
   $$
   \sin x = \cos 13^\circ.
   $$
3. Write $$\cos 13^\circ$$ as a sine:
   $$
   \cos 13^\circ = \sin(90^\circ - 13^\circ) = \sin 77^\circ.
   $$
   So:
   $$
   \sin x = \sin 77^\circ.
   $$
4. The general solution for $$\sin x = \sin \alpha$$ is:
   $$
   x = \alpha + 360^\circ k \quad \text{or} \quad x = 180^\circ - \alpha + 360^\circ k,\quad k\in\mathbb{Z}.
   $$
   With $$\alpha = 77^\circ$$, the solutions are:
   $$
   x = 77^\circ + 360^\circ k \quad \text{or} \quad x = 180^\circ - 77^\circ + 360^\circ k = 103^\circ + 360^\circ k.
   $$

**Final solutions for (5.6):**
$$
x = 77^\circ + 360^\circ k \quad \text{or} \quad x = 103^\circ + 360^\circ k,\quad k\in\mathbb{Z}.
$$
**(5.5) Solutions for $$ \sin 3x = \cos x $$:** $$ x = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{\pi}{4} + k\pi,\quad k\in\mathbb{Z}. $$ **(5.6) Solutions for $$ \sin x = \sin 43^\circ + \sin 17^\circ $$:** $$ x = 77^\circ + 360^\circ k \quad \text{or} \quad x = 103^\circ + 360^\circ k,\quad k\in\mathbb{Z}. $$

\( (5.5) \sin 3 x=\cos x \) (5.6) \( \sin x=\sin 43^{\circ}+\sin 17^{\circ} \)

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