Trig Functions — Question 1

Question 1 - Solution

(a) Identify parameters from $f(x) = 3\sin\!\left(2x - \frac{\pi}{3}\right) + 1$:

• Amplitude: $\boxed{3}$

• Period: $\dfrac{2\pi}{2} = \boxed{\pi}$

• Phase shift: $$2x - \frac{\pi}{3} = 0 \Rightarrow x = \frac{\pi}{6} \Rightarrow \boxed{\frac{\pi}{6} \text{ to the right}}$$

• Vertical shift: $\boxed{+1}$

(b) Maximum and Minimum:

Since the sine function oscillates between $-1$ and $1$: $$\text{Min of } f(x) = 3(-1) + 1 = -2, \quad \text{Max of } f(x) = 3(1) + 1 = 4$$ $$\boxed{\text{Max} = 4}, \quad \boxed{\text{Min} = -2}$$

(c) Sketch one full period:

One full period has length $\pi$, starting at the phase shift: $$x = \frac{\pi}{6} \quad \text{to} \quad x = \frac{7\pi}{6}$$

Key points over one period: $$\begin{array}{c|c} x & f(x) \\ \hline \frac{\pi}{6} & 1 \ (\text{midline}) \\ \frac{5\pi}{12} & 4 \ (\text{maximum}) \\ \frac{2\pi}{3} & 1 \ (\text{midline}) \\ \frac{11\pi}{12} & -2 \ (\text{minimum}) \\ \frac{7\pi}{6} & 1 \ (\text{midline}) \end{array}$$

(d) Solve $f(x) = 1$ on $[0, 2\pi]$:

$$3\sin\!\left(2x - \frac{\pi}{3}\right) + 1 = 1 \Rightarrow \sin\!\left(2x - \frac{\pi}{3}\right) = 0$$ $$2x - \frac{\pi}{3} = n\pi \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{6}$$

Values in $[0, 2\pi]$: $$\boxed{ x = \frac{\pi}{6},\ \frac{2\pi}{3},\ \frac{7\pi}{6},\ \frac{5\pi}{3} }$$