Chúng ta sẽ giải từng phương trình một cách chi tiết.

### Bài 26:

a) \(\sin\left(x - \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2}\)

Ta có \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), do đó:
\[ x - \frac{\pi}{6} = \frac{\pi}{4} + 2k\pi \quad \text{hoặc} \quad x - \frac{\pi}{6} = \pi - \frac{\pi}{4} + 2k\pi \]
\[ x - \frac{\pi}{6} = \frac{\pi}{4} + 2k\pi \quad \text{hoặc} \quad x - \frac{\pi}{6} = \frac{3\pi}{4} + 2k\pi \]
\[ x = \frac{\pi}{4} + \frac{\pi}{6} + 2k\pi \quad \text{hoặc} \quad x = \frac{3\pi}{4} + \frac{\pi}{6} + 2k\pi \]
\[ x = \frac{5\pi}{12} + 2k\pi \quad \text{hoặc} \quad x = \frac{11\pi}{12} + 2k\pi \]

b) \(\cos\left(x - \frac{\pi}{6}\right) = \frac{1}{2}\) với \(x \in (-\pi; \pi)\)

Ta có \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), do đó:
\[ x - \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi \quad \text{hoặc} \quad x - \frac{\pi}{6} = -\frac{\pi}{3} + 2k\pi \]
\[ x = \frac{\pi}{3} + \frac{\pi}{6} + 2k\pi \quad \text{hoặc} \quad x = -\frac{\pi}{3} + \frac{\pi}{6} + 2k\pi \]
\[ x = \frac{\pi}{2} + 2k\pi \quad \text{hoặc} \quad x = -\frac{\pi}{6} + 2k\pi \]

Với \(x \in (-\pi; \pi)\), ta có:
\[ x = \frac{\pi}{2} \quad \text{hoặc} \quad x = -\frac{\pi}{6} \]

c) \(\sin\left(x + \frac{\pi}{6}\right) = -\frac{1}{3}\)

Ta có \(\sin\left(\theta\right) = -\frac{1}{3}\), do đó:
\[ x + \frac{\pi}{6} = \arcsin\left(-\frac{1}{3}\right) + 2k\pi \quad \text{hoặc} \quad x + \frac{\pi}{6} = \pi - \arcsin\left(-\frac{1}{3}\right) + 2k\pi \]
\[ x = \arcsin\left(-\frac{1}{3}\right) - \frac{\pi}{6} + 2k\pi \quad \text{hoặc} \quad x = \pi - \arcsin\left(-\frac{1}{3}\right) - \frac{\pi}{6} + 2k\pi \]

d) \(\cos\left(-x + 30^\circ\right) = -\frac{\sqrt{2}}{2}\)

Ta có \(\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\), do đó:
\[ -x + 30^\circ = \frac{3\pi}{4} + 2k\pi \quad \text{hoặc} \quad -x + 30^\circ = -\frac{3\pi}{4} + 2k\pi \]
\[ -x = \frac{3\pi}{4} - 30^\circ + 2k\pi \quad \text{hoặc} \quad -x = -\frac{3\pi}{4} - 30^\circ + 2k\pi \]
\[ x = -\frac{3\pi}{4} + 30^\circ - 2k\pi \quad \text{hoặc} \quad x = \frac{3\pi}{4} + 30^\circ - 2k\pi \]

e) \(\sin 3x = \frac{\sqrt{3}}{2}\)

Ta có \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), do đó:
\[ 3x = \frac{\pi}{3} + 2k\pi \quad \text{hoặc} \quad 3x = \pi - \frac{\pi}{3} + 2k\pi \]
\[ 3x = \frac{\pi}{3} + 2k\pi \quad \text{hoặc} \quad 3x = \frac{2\pi}{3} + 2k\pi \]
\[ x = \frac{\pi}{9} + \frac{2k\pi}{3} \quad \text{hoặc} \quad x = \frac{2\pi}{9} + \frac{2k\pi}{3} \]

f) \(\sin^2\left(4x - \frac{7\pi}{3}\right) = \frac{3}{4}\)

Ta có \(\sin\left(\theta\right) = \pm\frac{\sqrt{3}}{2}\), do đó:
\[ 4x - \frac{7\pi}{3} = \frac{\pi}{3} + 2k\pi \quad \text{hoặc} \quad 4x - \frac{7\pi}{3} = \pi - \frac{\pi}{3} + 2k\pi \]
\[ 4x - \frac{7\pi}{3} = \frac{\pi}{3} + 2k\pi \quad \text{hoặc} \quad 4x - \frac{7\pi}{3} = \frac{2\pi}{3} + 2k\pi \]
\[ 4x = \frac{8\pi}{3} + 2k\pi \quad \text{hoặc} \quad 4x = \frac{9\pi}{3} + 2k\pi \]
\[ x = \frac{2\pi}{3} + \frac{k\pi}{2} \quad \text{hoặc} \quad x = \frac{3\pi}{4} + \frac{k\pi}{2} \]

### Bài 27:

a) \(\cot\left(x^2\right) = \sqrt{3}\)

Ta có \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\), do đó:
\[ x^2 = \frac{\pi}{6} + k\pi \]
\[ x = \pm\sqrt{\frac{\pi}{6} + k\pi} \]

b) \(\tan\left(x^2 + 4x + 2\right) = \tan 6\)

Ta có:
\[ x^2 + 4x + 2 = 6 + k\pi \]
\[ x^2 + 4x - 4 = k\pi \]
Giải phương trình bậc hai:
\[ x = \frac{-4 \pm \sqrt{16 + 4k\pi}}{2} \]
\[ x = -2 \pm \sqrt{4 + k\pi} \]

c) \(\tan\left(\pi(\cos x)\right) = \frac{\sqrt{3}}{3}\)

Ta có \(\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}\), do đó:
\[ \pi(\cos x) = \frac{\pi}{6} + k\pi \]
\[ \cos x = \frac{1}{6} + k \]

Hy vọng các bước giải này sẽ giúp bạn hiểu rõ hơn về cách giải các phương trình trên.