Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.

Using a trigonometric substituition:
As $-1 \le x \le 1$, then $x = \cos \theta$:
$$\sqrt{1+x} = \sqrt{1-\cos \theta} = \sqrt{2\sin^2 \frac{\theta}{2}} = \sqrt{2}\left|\sin \frac{\theta}{2}\right|$$
$$\sqrt{1+x} = \sqrt{1+\cos \theta} = \sqrt{2\cos^2 \frac{\theta}{2}} = \sqrt{2}\left|\cos \frac{\theta}{2}\right|$$
Then the equation becomes
$$\left|\sin \frac{\theta}{2}\right| + \left|\cos \frac{\theta}{2}\right| = \frac{1}{\sqrt{2}}$$
As we care only about the value of $x$, we can restrict the value of $\theta$ to be at the interval $\left[0, \ \pi\right]$:

*

*If $\theta \in \left[0, \ \dfrac{\pi}{2}\right]$:


$$\sin \frac{\theta}{2} + \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sqrt{2} \sin \underbrace{\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{\sqrt{2}}$$
$$\sin \alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3} \ \ \text{or} \ \ \alpha = \dfrac{2\pi}{3}$$
Then $\theta = 2\left(\alpha - \frac{\pi}{4}\right)$ can assume two values: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$.
Only $\theta = \frac{\pi}{6}$ is on the interval $\left[0, \ \frac{\pi}{2}\right]$, then we exclude the second value. Therefore:
$$\boxed{x = \cos \frac{\pi}{6} = \dfrac{\sqrt{3}}{2}}$$


*

*If $\theta \in \left[\dfrac{\pi}{2}, \ \pi\right]$

$$\sin \frac{\theta}{2} - \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sin \underbrace{\left(\dfrac{\theta}{2} - \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{2}$$
$$\alpha = \frac{\pi}{3} \ \ \text{or}  \ \  \alpha =  \frac{2 \pi}{3}$$
$$\theta = \frac{7\pi}{6} \ \ \text{or}  \ \  \theta =  \frac{11 \pi}{6} $$
As both $\theta$ are outside the interval $\left[\dfrac{\pi}{2}, \ \pi\right]$, there's no solution for this interval.

Then we got only one solution:
$$\boxed{x = \dfrac{\sqrt{3}}{2}}$$
