Solve $ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$. Is my solution correct?

Question

Find the roots of the following equation, if any: $$ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}. $$

My approach:

The following constraints should hold jointly for x:

• $1-x^2\geq0\iff x\in[-1,1]$
• $1+x\geq0 \iff x\geq-1$
• $1-x\geq0 \iff x\leq1$
• $\sqrt{1+x}\ne\sqrt{1-x}\iff x\ne0$

Consequently, $x\in[-1,0)\cup(0,1]$ and for such x's I solve and I reach at the following equation $$ \left(\frac{x}{1+\sqrt{1-x^2}}\right)^3=27\implies \frac{x}{1+\sqrt{1-x^2}}=3, $$ which gives $$ x = 3 + 3\sqrt{1-x^2}\implies 3\sqrt{1-x^2} = x-3\implies 9-9x^2=x^2-6x+9\implies 10x^2-6x=0\implies x(10x-6)=0, $$

and thus $x=0$ or $x=\frac{3}{5}$. The first one is rejected, but $x=\frac{3}{5}\in[-1,0)\cup(0,1]$.

However, if we plug $x=\frac{3}{5}$ into the original equation, we find out that this is impossible. So we should reject $x=\frac{3}{5}$. The original equation is impossible.

Is my approach correct? Thank you very much in advance.

Answer 1

$ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} $ is of the form $\frac{1-a}{1+a}$, which is $\leq1$.

$ 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} $ is of the form $27\frac{a+b}{a-b}$, which is $\geq27$.

So there is no solution.

Answer 2

The left-hand side is less than 1; the right-hand side is more than 27.

Answer 3

Let $2y=\arccos x\implies0\le2y\le\pi$

$\implies x=\cos2y$ and $\sqrt{1-x^2}=|\sin2y|=+\sin2y$ as $0\le2y\le\pi$

So, we have $$\dfrac{1-\sin2y}{1+\sin2y}=27\cdot\dfrac{\cos y+\sin y}{\cos y-\sin y}$$

$$\left(\dfrac{1-\tan y}{1+\tan y}\right)^2=27\cdot\dfrac{1+\tan y}{1-\tan y}$$

$$\iff\left(\dfrac{1-\tan y}{1+\tan y}\right)^3=27$$

As $\tan y$ is real, $$\dfrac{1-\tan y}{1+\tan y}=3\iff\tan y=-\dfrac12$$

But $\tan y\ge0$ as $0\le y\le\dfrac\pi2$