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How would I go about solving the equation below:

$\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$

After appyling sin to both sides I end up with:

$(2x^2-1)(\sqrt{1-x^2}^2 - x^2) + 2x \sqrt{1 -(2x^2-1)^2}\sqrt{1-x^2}= -1$

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    $\begingroup$ Have you tried applying $\sin$ to both sides? $\endgroup$ – Gibbs Sep 27 '18 at 9:21
  • $\begingroup$ I have and that only gives me the answers -1 and 0, however the answer is -1 <= x <= 0 $\endgroup$ – KGunnar Sep 27 '18 at 9:25
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    $\begingroup$ I suggest you show what you have tried. $\endgroup$ – Gibbs Sep 27 '18 at 9:26
  • $\begingroup$ I end up with this equation: (2x^2-1)(sqrt(1-x^2)^2 - x^2) + 2x* sqrt(1 -(2x^2-1)^2)(sqrt(1-x^2)) = -1 $\endgroup$ – KGunnar Sep 27 '18 at 9:30
  • $\begingroup$ Please, update your question with what you have tried. In the comments, especially without using MathJax, that equation is hard to read. $\endgroup$ – Gibbs Sep 27 '18 at 9:31
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Hint:

As $\arcsin(-y)=-\arcsin y$

$$\iff-2\arcsin x=\dfrac\pi2-\arcsin(1-2x^2)=\arccos(1-2x^2)$$

Now $0\le\arccos(1-2x^2)\le\pi\implies0\le-2\arcsin x\le\pi\iff0\ge\arcsin x\ge-\dfrac\pi2$

Now let $\arcsin x=y\implies x=\sin y,0\le-2y\le\pi$

$$\arccos(1-2x^2)=\arccos(\cos2y)=\begin{cases}2y &\mbox{if }0\le2y\le\pi\\ -2y& \mbox{if } 0\le-2y\le\pi \end{cases}$$

Can you take it from here?

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  • $\begingroup$ I do know there is not an easier than yours for this one! + $\endgroup$ – mrs Sep 27 '18 at 9:31
  • $\begingroup$ I’m having a hard time following this sorry. $\endgroup$ – KGunnar Sep 27 '18 at 11:10
  • $\begingroup$ @KGunnar, Please pinpoint your confusion $\endgroup$ – lab bhattacharjee Sep 27 '18 at 11:51
  • $\begingroup$ @KGunnar, Hope your are aware of en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Sep 27 '18 at 11:52
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A method using derivatives: put $f(x) := \arcsin (2x^2-1)+2\arcsin x$. Observe that $f$ is defined on $[-1,1]$, it is continuous and $f(0) = -\pi/2$. Let us compute its derivative: \begin{align} f'(x) & = \frac{4x}{\sqrt{1-(2x^2-1)^2}}+\frac{2}{\sqrt{1-x^2}} \\ & = \frac{2x}{\lvert x \rvert \sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}= \begin{cases} 0, \text{ if } -1 < x < 0 \\ \frac{4}{\sqrt{1-x^2}} \text{ if } 0 < x < 1. \end{cases} \end{align} So $f$ is constantly $-\pi/2$ on $[-1,0]$. For $x > 0$ its derivative is positive, so $f$ increases strictly and you do not get other solutions to your equation.

A remark on your method: if you decide to apply $\sin$ to both sides be aware that you are actually solving more than one equation, because $-1 = \sin \left(-\pi/2 + k\pi\right)$, with $k \in \mathbb{Z}$. So when you get the final solutions of $\sin(\arcsin(2x^2-1)+2\arcsin x)=-1$ you should check which ones are also solutions of, for example, $\arcsin (2x^2-1)+2\arcsin x = 3\pi/2$. You see, $x=1$ satisfies the latter equation, but not the former.

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The straight-forward method is to apply $\sin$ on both sides, the sine addition theorem, and to use the identity $\cos(\arcsin(\alpha)) = \sqrt{1-\alpha^2}$. This yields the following equation:

$$ -x^2(2x^2-1)+(2x^2-1)(1-x^2)+2x\sqrt{[1-(2x^2-1)^2][1-x^2]} = -1 \;. $$

Now, I just very quickly scribbled this on a piece of paper, so please double check it. After some term manipulations, I got $x\in\{-1,0,1\}$ as solutions.

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  • $\begingroup$ 1 is not a solution, however 0 and -1 are. All the solutions are within the interval -1 <= x <= 0 $\endgroup$ – KGunnar Sep 27 '18 at 9:39
  • $\begingroup$ The only non-equivalent operations I performed were sines and taking squares. This means that not all solutions of the modified equation are solutions to the original one, however, all solutions of the original one are contained in the set of solutions of the modified equation. In the example, $x=1$ is a solution only modulo periodicity: Plugging $x=1$ into your equation yields $\arcsin(1)+2\arcsin(1) = 3\pi/2 \simeq -\pi/2$. $\endgroup$ – Marc Mingoulis Sep 27 '18 at 9:53

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