which values are two of the possible solutions to the equation?\n\\( \\sin ( 5 x - \\pi ) = \\frac { \\sqrt…

Explanation:

Step1: Recall the values of sine function

We know that (\sin\theta=\frac{\sqrt{2}}{2}) when (\theta = \frac{\pi}{4}+ 2k\pi) or (\theta=\frac{3\pi}{4}+2k\pi), (k\in\mathbb{Z}).

So for (\sin(5x-\pi)=\frac{\sqrt{2}}{2}), we have (5x-\pi=\frac{\pi}{4}+2k\pi) or (5x - \pi=\frac{3\pi}{4}+2k\pi)

Step2: Solve for (x) in the first case

For (5x-\pi=\frac{\pi}{4}+2k\pi) [ \begin{align*} 5x&=\frac{\pi}{4}+\pi + 2k\pi\ 5x&=\frac{\pi + 4\pi}{4}+2k\pi\ 5x&=\frac{5\pi}{4}+2k\pi\ x&=\frac{\pi}{4}+\frac{2k\pi}{5} \end{align*} ]

Step3: Solve for (x) in the second case

For (5x-\pi=\frac{3\pi}{4}+2k\pi) [ \begin{align*} 5x&=\frac{3\pi}{4}+\pi+2k\pi\ 5x&=\frac{3\pi + 4\pi}{4}+2k\pi\ 5x&=\frac{7\pi}{4}+2k\pi\ x&=\frac{7\pi}{20}+\frac{2k\pi}{5} \end{align*} ]

Step4: Find values of (x) in the range (0\leq x<2\pi)

When (k = 0) in (x=\frac{\pi}{4}+\frac{2k\pi}{5}), (x=\frac{\pi}{4}) When (k = 0) in (x=\frac{7\pi}{20}+\frac{2k\pi}{5}), (x=\frac{7\pi}{20})

Answer:

(\left{\frac{\pi}{4},\frac{7\pi}{20}\right})