1. Solve $\log_2(3x + 1) - 2\log_2 x = \log_{\frac{1}{3}} 27$. Step 1: Use log properties: $2\log_2 x = \log_2 x^2$. Step 2: Rewrite equation as $\log_2(3x + 1) - \log_2 x^2 = \log_{\frac{1}{3}} 27$. Step 3: Use subtraction property: $\log_2 \frac{3x + 1}{x^2} = \log_{\frac{1}{3}} 27$. Step 4: Convert right side: $\log_{\frac{1}{3}} 27 = y$ means $(\frac{1}{3})^y = 27$. Step 5: Express $27 = 3^3$, so $(\frac{1}{3})^y = 3^3$. Step 6: Rewrite $(\frac{1}{3})^y = 3^{-y}$, so $3^{-y} = 3^3$ implies $-y = 3$ or $y = -3$. Step 7: So $\log_2 \frac{3x + 1}{x^2} = -3$. Step 8: Rewrite as $\frac{3x + 1}{x^2} = 2^{-3} = \frac{1}{8}$. Step 9: Multiply both sides by $x^2$: $3x + 1 = \frac{x^2}{8}$. Step 10: Multiply both sides by 8: $8(3x + 1) = x^2$ or $24x + 8 = x^2$. Step 11: Rearrange: $x^2 - 24x - 8 = 0$. Step 12: Use quadratic formula: $x = \frac{24 \pm \sqrt{24^2 + 4 \times 8}}{2} = \frac{24 \pm \sqrt{576 + 32}}{2} = \frac{24 \pm \sqrt{608}}{2}$. Step 13: Simplify $\sqrt{608} = \sqrt{16 \times 38} = 4\sqrt{38}$. Step 14: So $x = \frac{24 \pm 4\sqrt{38}}{2} = 12 \pm 2\sqrt{38}$. Step 15: Check domain: $x > 0$ and $3x + 1 > 0$. Step 16: Both solutions positive, so both valid. Answer 1: $x = 12 + 2\sqrt{38}$ or $x = 12 - 2\sqrt{38}$. 2. Solve $\left(\frac{1}{27}\right)^{3x} 9^{x-1} = \log_2 8$. Step 1: Express bases as powers of 3: $\frac{1}{27} = 3^{-3}$, $9 = 3^2$, and $\log_2 8 = 3$. Step 2: Rewrite: $(3^{-3})^{3x} (3^2)^{x-1} = 3$. Step 3: Simplify exponents: $3^{-9x} \times 3^{2x - 2} = 3$. Step 4: Combine powers: $3^{-9x + 2x - 2} = 3$ or $3^{-7x - 2} = 3^1$. Step 5: Equate exponents: $-7x - 2 = 1$. Step 6: Solve for $x$: $-7x = 3$ so $x = -\frac{3}{7}$. Answer 2: $x = -\frac{3}{7}$. 3. Solve $\cos 2x - \cos x + \cos^2 x + \sin^2 x = 0$. Step 1: Use identity $\cos^2 x + \sin^2 x = 1$. Step 2: Substitute: $\cos 2x - \cos x + 1 = 0$. Step 3: Use double angle formula: $\cos 2x = 2\cos^2 x - 1$. Step 4: Substitute: $2\cos^2 x - 1 - \cos x + 1 = 0$. Step 5: Simplify: $2\cos^2 x - \cos x = 0$. Step 6: Factor: $\cos x (2\cos x - 1) = 0$. Step 7: Set each factor to zero: - $\cos x = 0$ gives $x = \frac{\pi}{2} + k\pi$. - $2\cos x - 1 = 0$ gives $\cos x = \frac{1}{2}$, so $x = \pm \frac{\pi}{3} + 2k\pi$. Answer 3: $x = \frac{\pi}{2} + k\pi$ or $x = \pm \frac{\pi}{3} + 2k\pi$, $k \in \mathbb{Z}$. 4. Solve $\sin x + \cos x = \sqrt{0.5}$. Step 1: Note $\sqrt{0.5} = \frac{\sqrt{2}}{2}$. Step 2: Use identity: $\sin x + \cos x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right)$. Step 3: Set $\sqrt{2} \sin \left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$. Step 4: Divide both sides by $\sqrt{2}$: $\sin \left(x + \frac{\pi}{4}\right) = \frac{1}{2}$. Step 5: Solve for $x + \frac{\pi}{4}$: $x + \frac{\pi}{4} = \frac{\pi}{6} + 2k\pi$ or $x + \frac{\pi}{4} = \frac{5\pi}{6} + 2k\pi$. Step 6: Subtract $\frac{\pi}{4}$: $x = \frac{\pi}{6} - \frac{\pi}{4} + 2k\pi = -\frac{\pi}{12} + 2k\pi$ or $x = \frac{5\pi}{6} - \frac{\pi}{4} + 2k\pi = \frac{7\pi}{12} + 2k\pi$. Answer 4: $x = -\frac{\pi}{12} + 2k\pi$ or $x = \frac{7\pi}{12} + 2k\pi$, $k \in \mathbb{Z}$. 5. Solve $\sqrt{(3 - x)^2} = 5^{\log_5 9}$. Step 1: Simplify left side: $\sqrt{(3 - x)^2} = |3 - x|$. Step 2: Simplify right side: $5^{\log_5 9} = 9$. Step 3: Equation: $|3 - x| = 9$. Step 4: Solve absolute value: - $3 - x = 9$ gives $x = -6$. - $3 - x = -9$ gives $x = 12$. Answer 5: $x = -6$ or $x = 12$. 6. Solve $[2 - 0.5x] = -3$. Step 1: Assuming $[\cdot]$ is the floor function. Step 2: $\lfloor 2 - 0.5x \rfloor = -3$ means $-3 \leq 2 - 0.5x < -2$. Step 3: Solve inequalities: From $-3 \leq 2 - 0.5x$: $-5 \leq -0.5x$ so $10 \geq x$. From $2 - 0.5x < -2$: $-0.5x < -4$ so $x > 8$. Step 4: Combine: $8 < x \leq 10$. Answer 6: $x \in (8, 10]$. Final answers: 1) $x = 12 \pm 2\sqrt{38}$ 2) $x = -\frac{3}{7}$ 3) $x = \frac{\pi}{2} + k\pi$ or $x = \pm \frac{\pi}{3} + 2k\pi$ 4) $x = -\frac{\pi}{12} + 2k\pi$ or $x = \frac{7\pi}{12} + 2k\pi$ 5) $x = -6$ or $x = 12$ 6) $x \in (8, 10]$