(Should be) simple trigonometric equation: \(\displaystyle \sin3x=\sin6x\).
(My) reasoning: the solution is either [1] \(\displaystyle \sin=0\) or [2] \(\displaystyle \sin3x\) and \(\displaystyle \sin6x\) are in different quadrants (I. and II. OR III. and IV. quadrant). For [1], \(\displaystyle \sin3x=0 \rightarrow 3x=k\pi \rightarrow x=\frac{k\pi}{3}\). For [2], (I'd say
) \(\displaystyle 3x=\pi-6x \Rightarrow 9x=\pi+2k\pi \rightarrow x=\frac{\pi}{9}+\frac{2k\pi}{9}\) (for both, [1] and [2], \(\displaystyle k\in\mathbb{I}\) (whole numbers))
Solutions given are these three (not two! \(\displaystyle \rightarrow\) how should I formalize \(\displaystyle 3x=\pi-6x\) to include, not only I. and II. quadrant, but also III. and IV. quadrant?): [1] \(\displaystyle x=\frac{k\pi}{3}\), [2a] \(\displaystyle x=\frac{\pi}{9}+\frac{2k\pi}{3}\) (why do I get \(\displaystyle \frac{2k\pi}{9}\)?!), [2b] \(\displaystyle x=-\frac{\pi}{9}+\frac{2k\pi}{3}\) (again, how to include the second "pair" of quadrants?).
My logic in such cases, when I'm asked to mathematically formalize the solution, isn't so shabby ... BUT ... I lack rigo(u)r to a always get the "strictly" correct result, and b not miss some solutions (for instance, [2a] solution (which I'm aware of when solving, but don't know how to formalize)).
AD: I should rather do \(\displaystyle \sin6x-\sin3x=0 \rightarrow 2\cos\frac{9x}{2}\sin\frac{3x}{2}=0\). Either \(\displaystyle \frac{3x}{2}=k\pi\) or \(\displaystyle \frac{9x}{2}=\frac{\pi}{2}+k\pi\) . But from \(\displaystyle \frac{3x}{2}=k\pi\) you get \(\displaystyle x=\frac{2k\pi}{3}\)! (and not \(\displaystyle x=\frac{k\pi}{3}\), which is given as solution). Could it be, that they've messed up the solutions, so that the [1] solution (\(\displaystyle x=\frac{2k\pi}{3}\)) is misput as part of [2a] and [2b] solutions (whose second parts should instead read as \(\displaystyle \frac{2k\pi}{9}\)) :evil: ?
(My) reasoning: the solution is either [1] \(\displaystyle \sin=0\) or [2] \(\displaystyle \sin3x\) and \(\displaystyle \sin6x\) are in different quadrants (I. and II. OR III. and IV. quadrant). For [1], \(\displaystyle \sin3x=0 \rightarrow 3x=k\pi \rightarrow x=\frac{k\pi}{3}\). For [2], (I'd say
) \(\displaystyle 3x=\pi-6x \Rightarrow 9x=\pi+2k\pi \rightarrow x=\frac{\pi}{9}+\frac{2k\pi}{9}\) (for both, [1] and [2], \(\displaystyle k\in\mathbb{I}\) (whole numbers))Solutions given are these three (not two! \(\displaystyle \rightarrow\) how should I formalize \(\displaystyle 3x=\pi-6x\) to include, not only I. and II. quadrant, but also III. and IV. quadrant?): [1] \(\displaystyle x=\frac{k\pi}{3}\), [2a] \(\displaystyle x=\frac{\pi}{9}+\frac{2k\pi}{3}\) (why do I get \(\displaystyle \frac{2k\pi}{9}\)?!), [2b] \(\displaystyle x=-\frac{\pi}{9}+\frac{2k\pi}{3}\) (again, how to include the second "pair" of quadrants?).
My logic in such cases, when I'm asked to mathematically formalize the solution, isn't so shabby ... BUT ... I lack rigo(u)r to a always get the "strictly" correct result, and b not miss some solutions (for instance, [2a] solution (which I'm aware of when solving, but don't know how to formalize)).
AD: I should rather do \(\displaystyle \sin6x-\sin3x=0 \rightarrow 2\cos\frac{9x}{2}\sin\frac{3x}{2}=0\). Either \(\displaystyle \frac{3x}{2}=k\pi\) or \(\displaystyle \frac{9x}{2}=\frac{\pi}{2}+k\pi\) . But from \(\displaystyle \frac{3x}{2}=k\pi\) you get \(\displaystyle x=\frac{2k\pi}{3}\)! (and not \(\displaystyle x=\frac{k\pi}{3}\), which is given as solution). Could it be, that they've messed up the solutions, so that the [1] solution (\(\displaystyle x=\frac{2k\pi}{3}\)) is misput as part of [2a] and [2b] solutions (whose second parts should instead read as \(\displaystyle \frac{2k\pi}{9}\)) :evil: ?
