Solve the equation: $\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$

Question

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:

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

Answer 1

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?

Answer 2

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.

Answer 3

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.