Question Type 4: Applying transformation to trigonometric functions under a specific domain Exercises

Given $y = -\sin\left(2x + \frac{\pi}{3}\right)$, find one interval of length one period within $-\pi < x < \pi$ and list all $x$ in that interval where $y=0$.

For $y = \sin\left(3x - \frac{\pi}{2}\right)$, find the phase shift and the first positive $x$-intercept in $-\pi < x < \pi$.

Identify the amplitude and period of $y = 2\sin\left(\frac{x}{3}\right)$ on the domain $-\pi < x < \pi$.

Find the period and midline of $y = 4\sin\bigl(\frac{x}{4}\bigr) - 3$.

Solve $\sin\bigl(\frac{x}{2}\bigr) + 2 = 0$ for $-\pi < x < \pi$.

For $y = 5\sin(3x - \pi) + 2$, find the amplitude, period, and equation of the principal axis.

Solve the equation $2\sin\bigl(\frac{x}{3}\bigr) - 1 = 0$ for $-\pi < x < \pi$.

Determine the maximum and minimum values of $y = 3\cos\bigl(2x + \pi\bigr)$ on $-\pi < x < \pi$.

Determine the range and period of $y = -2\cos\bigl(\frac{x}{6}\bigr) + 4$.

Write the function obtained by shifting $y = 2\sin\left(\frac{x}{3}\right)$ up by 1 unit, and determine its new maximum value.

Find all $x$-intercepts of $y = 2\sin\bigl(\frac{x}{3}\bigr)$ in the interval $-\pi < x < \pi$.

Solve $5\sin\bigl(3x - \pi\bigr) + 2 = 2$ for $-\pi < x < \pi$.