Solve arcsin(x) + arcsin(2x) = pi/2.

Step-by-step solution

1. 1

   Establish the domain first. Both inverse sines must be defined, so $-1\le x\le1$ and $-1\le2x\le1$. The second is the stricter condition, giving

   $-\tfrac12\le x\le\tfrac12.$

   Any candidate outside this interval is automatically invalid, no matter what the algebra produces.

2. 2

   Isolate one inverse sine and take the sine of both sides. Rearranging and applying $\sin$:

   $\arcsin(2x)=\frac{\pi}{2}-\arcsin(x)\;\Longrightarrow\;2x=\sin\!\left(\frac{\pi}{2}-\arcsin x\right)=\cos(\arcsin x).$

   The co-function identity $\sin\!\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$ is what makes the right side manageable.

3. 3

   Convert $\cos(\arcsin x)$ into algebra. Put $\theta=\arcsin x$, so $\sin\theta=x$ and $\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ — an interval on which $\cos\theta\ge0$. Hence the positive root is taken:

   $\cos(\arcsin x)=\sqrt{1-x^{2}},$

   and the equation becomes $2x=\sqrt{1-x^{2}}$.

4. 4

   Extract the sign condition before squaring. A square root is never negative, so $2x\ge0$, i.e. $x\ge0$. Recording this now is what prevents an extraneous root from surviving the squaring step. Squaring:

   $4x^{2}=1-x^{2}\;\Longrightarrow\;5x^{2}=1\;\Longrightarrow\;x=\pm\frac{1}{\sqrt5}.$

5. 5

   Select and verify the root. The condition $x\ge0$ eliminates the negative value, leaving

   $x=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}\approx0.4472,$

   which also lies inside the domain $\left[-\tfrac12,\tfrac12\right]$ ✓. Checking directly: $\arcsin(0.4472)=0.46365$ and $\arcsin(0.8944)=1.10715$, and $0.46365+1.10715=1.57080=\tfrac{\pi}{2}$ ✓. The rejected root $x=-\tfrac{1}{\sqrt5}$ gives $-\tfrac{\pi}{2}$ instead.