I have a few set of these questions which I'm kind of confused about how to answer. I don't particularly need an answer just an explanation of how it's done.
Provide answers in the form $\frac pq \pi$, with $\frac pq$ a fraction in the lowest possible terms.
Find $x$, $-\frac 12 \pi \le x \le \frac 12 \pi$, such as that $\sin x = \sin\left(\frac {97}{17} \pi\right)$.
I don't quite understand what is meant by the less than $x$, this question is maybe really simple as it's early on in my worksheet but I just have no idea on how to address it. Thanks
edit: grammar
Consider the diagram below:
Two angles have the same sine if the $y$-coordinates of the points where they intersect the unit circle are equal. Hence, $\sin\theta = \sin\varphi$ if $\varphi = \theta$ or $\varphi = \pi - \theta$. Any angle coterminal with these angles will also have the same sine. In general, $\sin\theta = \sin\varphi$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2m\pi, m \in \mathbb{Z}$$
You wish to solve the equation $$\sin x = \sin\left(\frac{97}{17}\pi\right)$$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. The general solution to the equation is $$x = \frac{97}{17}\pi + 2k\pi, k \in \mathbb{Z}$$ or \begin{align*} x & = \pi - \frac{97}{17}\pi + 2m\pi, m \in \mathbb{Z}\\ & = -\frac{80}{17}\pi + 2m\pi, m \in \mathbb{Z} \end{align*} Observe that if we set $k = -3$, we obtain $$x = \frac{97}{17}\pi - 6\pi = -\frac{5}{17}\pi$$ and that $$-\frac{\pi}{2} \leq -\frac{5}{17}\pi \leq \frac{\pi}{2}$$ Since the angles satisfying the equation $$x = \frac{97}{17}\pi + 2k\pi, k \in \mathbb{Z}$$ differ by integer multiples of $2\pi$, any other value of $k$ will yield an angle outside the desired interval.
Observe that if $m = 2$, then $$x = -\frac{80}{17}\pi + 4\pi = -\frac{12}{17}\pi < -\frac{\pi}{2}$$ and that if $m = 3$, then $$x = -\frac{80}{17}\pi + 6\pi = \frac{22}{17}\pi > \frac{\pi}{2}$$ Since the angles satisfying the equation $$x = -\frac{80}{17}\pi + 2m\pi, m \in \mathbb{Z}$$ differ by integer multiples of $2\pi$, any other value of $m$ will also yield an angle outside the desired interval.
Hence, the only valid solution of the equation $\sin x = \sin\left(\frac{97}{17}\pi\right)$ satisfying the inequalities $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ is $x = -\frac{5\pi}{17}$.