1. **Problem:** Given the function $$f(x) = 3 \cos\left(\frac{\pi x}{6}\right) + 1$$ for $$0 \leq x \leq 12$$, answer the following: 2. **a. Amplitude:** The amplitude of a cosine function $$a \cos(bx) + d$$ is the absolute value of $$a$$. Here, $$a = 3$$, so the amplitude is $$3$$. 3. **b. Values of $$b$$ and $$d$$:** The function is $$f(x) = 3 \cos\left(\frac{\pi x}{6}\right) + 1$$. So, $$b = \frac{\pi}{6}$$ and $$d = 1$$. 4. **c. Period:** The period $$T$$ of $$\cos(bx)$$ is given by $$T = \frac{2\pi}{b}$$. Substitute $$b = \frac{\pi}{6}$$: $$ T = \frac{2\pi}{\frac{\pi}{6}} = 2\pi \times \frac{6}{\pi} = 12 $$ 5. **d. Sketching the function:** - The amplitude is 3, so the function oscillates between $$1 - 3 = -2$$ and $$1 + 3 = 4$$. - The period is 12, so one full cycle occurs from $$x=0$$ to $$x=12$$. - At $$x=0$$, $$f(0) = 3 \cos(0) + 1 = 3(1) + 1 = 4$$ (maximum). - At $$x=6$$ (half period), $$f(6) = 3 \cos\left(\frac{\pi \times 6}{6}\right) + 1 = 3 \cos(\pi) + 1 = 3(-1) + 1 = -2$$ (minimum). - At $$x=12$$ (full period), $$f(12) = 3 \cos(2\pi) + 1 = 3(1) + 1 = 4$$ (maximum). - The graph is a cosine wave shifted vertically by 1. --- 5. **Problem:** Express $$2\cos^2 x + \sin x - 1$$ in the form $$a \sin^2 x + b \sin x + c$$ and solve $$2\cos^2 x + \sin x - 1 = 0$$ for $$0 \leq x \leq 2\pi$$. 6. **a. Expression rewrite:** Use the identity $$\cos^2 x = 1 - \sin^2 x$$: $$ 2\cos^2 x + \sin x - 1 = 2(1 - \sin^2 x) + \sin x - 1 = 2 - 2\sin^2 x + \sin x - 1 = -2\sin^2 x + \sin x + 1 $$ Multiply by -1 to write in standard quadratic form: $$ -2\sin^2 x + \sin x + 1 = 0 \implies 2\sin^2 x - \sin x - 1 = 0 $$ So, $$a=2$$, $$b=-1$$, $$c=-1$$. 7. **b. Solve the equation:** Solve $$2\sin^2 x - \sin x - 1 = 0$$ for $$0 \leq x \leq 2\pi$$. Let $$u = \sin x$$, then: $$ 2u^2 - u - 1 = 0 $$ Use quadratic formula: $$ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} $$ So, $$ u_1 = \frac{1 + 3}{4} = 1, \quad u_2 = \frac{1 - 3}{4} = -\frac{1}{2} $$ For $$u_1 = 1$$: $$ \sin x = 1 \implies x = \frac{\pi}{2} $$ For $$u_2 = -\frac{1}{2}$$: $$ \sin x = -\frac{1}{2} \implies x = \frac{7\pi}{6}, \frac{11\pi}{6} $$ Therefore, solutions are: $$ x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} $$ --- 8. **Problem:** Express $$32 \sin x \cos x$$ in the form $$a \sin bx$$ and solve $$32 \sin x \cos x = 8$$ for $$0 \leq x \leq \pi$$. 9. **a. Expression rewrite:** Use the double-angle identity: $$ \sin 2x = 2 \sin x \cos x \implies 32 \sin x \cos x = 16 \times 2 \sin x \cos x = 16 \sin 2x $$ So, $$a = 16$$ and $$b = 2$$. 10. **b. Solve the equation:** $$ 32 \sin x \cos x = 8 \implies 16 \sin 2x = 8 \implies \sin 2x = \frac{8}{16} = \frac{1}{2} $$ Solve for $$2x$$: $$ \sin 2x = \frac{1}{2} \implies 2x = \frac{\pi}{6}, \frac{5\pi}{6} \quad \text{(within } 0 \leq 2x \leq 2\pi \text{)} $$ Divide by 2: $$ x = \frac{\pi}{12}, \frac{5\pi}{12} $$ Since $$x$$ must be in $$[0, \pi]$$, these are valid solutions. --- **Final answers:** - 1a: Amplitude = 3 - 1b: $$b = \frac{\pi}{6}$$, $$d = 1$$ - 1c: Period = 12 - 5a: $$2\cos^2 x + \sin x - 1 = 2\sin^2 x - \sin x - 1$$ - 5b: Solutions $$x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$$ - 6a: $$32 \sin x \cos x = 16 \sin 2x$$, so $$a=16$$, $$b=2$$ - 6b: Solutions $$x = \frac{\pi}{12}, \frac{5\pi}{12}$$