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Find the general solution of:

$$\sin^3 \theta - \sin \theta = 0$$

Working out: (Factorise out)

$$\sin \theta (\sin^2 \theta - 1) = 0$$

Solve for $\sin \theta$ and $\sin^2 \theta - 1$:

For $\sin \theta = 0$

$$\sin \theta = 0$$

$$\therefore \theta = 0$$

$$\therefore \theta = n \pi$$

For $\sin^2 \theta - 1 = 0$:

$$\sin^2 \theta - 1 = 0$$

$$\sin^2 \theta = 1$$

$$\sin \theta = \pm1$$

$$\theta = - \pi/2, \pi/2$$

$$\therefore \theta = n\pi + (-1)^n(\pm\pi/2)$$

So i get the two solutions as:

$$\theta = n\pi\text{ and }\theta = n\pi + (-1)^n(\pm\pi/2)$$

However a complication occurs because wolframalpha says otherwise, help would be greatly appreciated.

Wolframalpha

MATHSUSER
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  • Hint: $sin^2\theta - 1 = - cos^2\theta$ – Lost Dec 10 '13 at 04:30
  • I would write $n\pi +\frac{\pi}{2}$ for the $\sin^2\theta=1$ case. Or maybe $2n\pi \pm \frac{\pi}{2}$. You are entioning everybody twice. Which is not wrong, but it is not optimal. – André Nicolas Dec 10 '13 at 04:33
  • @AndréNicolas Can someone explain to me why my one is wrong though? – MATHSUSER Dec 10 '13 at 04:36
  • @Lost Why would I make it equal to $-\cos^2 \theta$?, why not just use what's provided? – MATHSUSER Dec 10 '13 at 04:37
  • Take $n=0$, and the $+$. We get $\frac{\pi}{2}$. Take $n=1$, and the $+$. We get $\frac{\pi}{2}$. Right, but stuttery. You should also not talk about $2$ solutions, it is $2$ (or $3$, depends how one counts) families of solutions. – André Nicolas Dec 10 '13 at 04:38
  • Because then you get $-cos^2\theta=0$ which would be easier to solve, and thus you'd have been able to compare with the answer you got. The way you did was fine, but it's easier to trip up in my opinion. – Lost Dec 10 '13 at 04:39
  • @MATHSUSER: I think I have explained. And it is not wrong, but your expression mentions each of the solutions of $\sin^2\theta=1$ twice. – André Nicolas Dec 10 '13 at 04:42
  • @Lost But i still get $\pi/2$? – MATHSUSER Dec 10 '13 at 04:42
  • @AndréNicolas - but that's because of the square root. But can't I just write them separately if that's the case like this?

    $n\pi + (-1)^n (\pi/2)$ and $n\pi + (-1)^n (-\pi/2)$

     – MATHSUSER Dec 10 '13 at 04:45
  • Precisely - that is correct. As André says, you aren't wrong, but could write the solution better. Changing the equation in terms of cosine hopefully can help you eliminate any confusion relating to $\pm 1$ – Lost Dec 10 '13 at 04:46
  • Oh okay thank you, so I am not wrong, but the only problem is me writing the $\pm 1$. I am just confused at how wolframalpha reaches that conclusion that is all. – MATHSUSER Dec 10 '13 at 04:48
  • This is OK, and mentions everybody only once. There are many ways one can describe the solutions. I would call $n\pi+\frac{\pi}{2}$, or the equivalent $(2n+1)\frac{\pi}{2}$ simpler-looking than your version. However, your revised version is correct, and non-repetitive. – André Nicolas Dec 10 '13 at 04:50
  • @AndréNicolas Alright thank you. – MATHSUSER Dec 10 '13 at 04:52

2 Answers2

1

$\displaystyle\cos^2\theta=0$ is definitely the easiest as it implies and is impled by $\displaystyle\cos\theta=0\implies \theta$ is odd multiple of $\displaystyle \frac\pi2$

Otherwise using Double-Angle Formulas,

$\displaystyle\sin^2\theta=\sin^2\alpha\iff \cos^2\theta=\cos^2\alpha$ $\displaystyle\iff\tan^2\theta=\tan^2\alpha$ $\displaystyle\iff\cos2\theta=\cos2\alpha$

$\displaystyle\implies 2\theta=2n\pi\pm2\alpha $ where $n$ is any integer

lab bhattacharjee
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When you write the solution for $$\ (\sin^2 \theta - 1) = 0$$ your answer is correct but you did not notice that what you wrote is exactly an odd multiple of $\displaystyle \frac\pi2$ as mentioned by lab bhattacharjee.

Claude Leibovici
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