Sine Period 05Bb6D

1. **State the problem:** We are given the function $$y=1-3\sin2\left(x+\frac{\pi}{3}\right)$$ and need to find its period, the interval for one period, the starting point of the interval, and 5 specific $x$-values within that interval. 2. **Recall the formula for the period of a sine function:** For a function of the form $$y = A \sin(B(x - C)) + D,$$ the period is given by $$\text{Period} = \frac{2\pi}{|B|}.$$ Here, $B$ is the coefficient of $x$ inside the sine. 3. **Identify $B$ in our function:** The function is $$y=1-3\sin2\left(x+\frac{\pi}{3}\right) = 1 - 3\sin\left(2x + \frac{2\pi}{3}\right).$$ So, $B = 2$. 4. **Calculate the period:** $$\text{Period} = \frac{2\pi}{|2|} = \frac{2\pi}{2} = \pi.$$ 5. **Find the starting point of the period interval:** The inside of the sine is $$2\left(x + \frac{\pi}{3}\right) = 2x + \frac{2\pi}{3}.$$ The sine function completes one period when its argument increases by $2\pi$. So, the interval for $x$ satisfies: $$2\left(x + \frac{\pi}{3}\right) \in \left[ -\frac{\pi}{2}, -\frac{\pi}{2} + 2\pi \right)$$ But usually, the period interval starts where the argument inside sine equals zero: $$2\left(x + \frac{\pi}{3}\right) = 0 \implies x + \frac{\pi}{3} = 0 \implies x = -\frac{\pi}{3}.$$ So the period interval starts at $$x = -\frac{\pi}{3}.$$ 6. **Write the period interval:** Since the period is $\pi$, the interval is: $$\left[-\frac{\pi}{3}, -\frac{\pi}{3} + \pi \right) = \left[-\frac{\pi}{3}, \frac{2\pi}{3}\right).$$ 7. **Find 5 $x$-values evenly spaced in this interval:** The interval length is $\pi$, so spacing 5 points evenly means spacing by $\frac{\pi}{4}$ (since 4 intervals between 5 points): - $x_1 = -\frac{\pi}{3}$ - $x_2 = -\frac{\pi}{3} + \frac{\pi}{4} = -\frac{4\pi}{12} + \frac{3\pi}{12} = -\frac{\pi}{12}$ - $x_3 = -\frac{\pi}{3} + 2\cdot\frac{\pi}{4} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}$ - $x_4 = -\frac{\pi}{3} + 3\cdot\frac{\pi}{4} = -\frac{\pi}{3} + \frac{3\pi}{4} = \frac{5\pi}{12}$ - $x_5 = -\frac{\pi}{3} + 4\cdot\frac{\pi}{4} = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$ **Final answers:** - Period: $$\pi$$ - Interval: $$\left[-\frac{\pi}{3}, \frac{2\pi}{3}\right)$$ - Starting point: $$-\frac{\pi}{3}$$ - Five $x$-values: $$-\frac{\pi}{3}, -\frac{\pi}{12}, \frac{\pi}{6}, \frac{5\pi}{12}, \frac{2\pi}{3}$$