I need help with this problem. I am aware that this problem has been posted before but the answers didn't really satisfy me.
This is how far I've gotten with this:
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1If $t=sin(x)$ and you have solved for $t$, then $x = \arcsin (t)$ for your $t$ values no? – IntegrateThis Nov 02 '21 at 02:16
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5Please link to the previous postings of the problem you say you have found, and explain what you find unsatisfying about their answers. – Troposphere Nov 02 '21 at 02:17
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I'm not sure how to help but, cool last line: $-\frac{\phi}{2}$! – rhkoulen Nov 02 '21 at 02:17
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@IntegrateThis As you can see I don’t know how to solve for x when sin x = -(1 +/- sqrt(5))/4 – leun Nov 02 '21 at 02:18
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@leun As they said, just take the inverse sin of both sides – rhkoulen Nov 02 '21 at 02:19
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@rhkoulen he is expecting some kind of ratio of $2 \pi$ I'm guessing. – IntegrateThis Nov 02 '21 at 02:20
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Why not use $\cos \theta = \sin ((4n+1) \pi / 2 - \theta)$ and similar? – Math Lover Nov 02 '21 at 02:24
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Ya, so try and find $a,b$ with $a+b = \frac{3\pi}{10}$ where $sin(a+b) = t$ and $sin(a)$ and $sin(b)$ have explicit solutions using special triangles/sin identities. – IntegrateThis Nov 02 '21 at 02:27
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@Troposphere he has three values for $t$, one being $1$ where $\arcsin(1) = \pi$ – IntegrateThis Nov 02 '21 at 02:30
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$3x+2x=\frac\pi2$ where $x=…$ – ACB Nov 02 '21 at 02:33
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@ACB That should be $\frac\pi2 + 2n\pi$ for an arbitrary integer $n,$ in order to capture all solutions. – David K Nov 02 '21 at 02:34
2 Answers
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Alternate approach:
We have
\begin{align*} \sin(3x)=\cos(2x)&\implies\sin(3x)=\sin(2x+\pi/2)\\ &\implies\sin(3x)-\sin(2x+\pi/2)=0\\ &\implies2\cos\left(\frac{3x+2x+\pi/2}{2}\right)\sin\left(\frac{3x-2x-\pi/2}{2}\right)=0\\ &\implies2\cos\left(\frac{5x+\pi/2}{2}\right)\sin\left(\frac{x-\pi/2}{2}\right)=0\\ \end{align*}
So,
\begin{align*} \cos\left(\frac{5x+\pi/2}{2}\right)=0&&\text{or}&&\sin\left(\frac{x-\pi/2} {2}\right)=0\\ \frac{5x+\pi/2}{2}=\frac{\pi}{2}+n\pi&&\text{or}&&\frac{x-\pi/2} {2}=n\pi&&\forall n\in\mathbb Z\\ x=\frac{4n+1}{10}\pi&&\text{or}&&x=\frac{4n+1}2\pi&&\forall n\in\mathbb Z\\ \end{align*}
Bonnaduck
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$n$ in the statements $T(x)=n\pi$ and $x=G(n)$ is quantified with $\exists$, not with $\forall$. – Nov 02 '21 at 06:48
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$\sin{3x}=\cos(\frac{\pi}{2}-3x)=\cos{2x}$ $\frac{\pi}{2}-3x=\pm{2x}+2k\pi$
where $k\in Z$
$x=\frac{\pi}{10}+\frac{2k\pi}{5}$
OR
$x=\frac{\pi}{2}+{2k\pi}$
Lion Heart
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