Miscellaneous · Q14 (MCQ): Solve $\sin^{-1}(1-x)-2\sin^{-1}x=\frac\pi2$

Basics you need

Solution

1. The domain: x must be in [0, 1]

   $\sin^{-1}x$ needs $-1 \le x \le 1$⁠. $\sin^{-1}(1-x)$ needs $-1 \le 1-x \le 1$⁠, that is $0 \le x \le 2$⁠. Both hold exactly when

   $$0 \le x \le 1.$$
   Only these $x$ can be solutions, because outside them one of the terms does not exist. This interval also controls the sizes of both terms in the next step.
   Domain of $\sin^{-1}$ is $[-1,1]$ (Basic 1).

2. Compare sizes: the equation forces x = 0

   Let $\alpha = \sin^{-1}x$ and $\beta = \sin^{-1}(1-x)$⁠. For $0\le x\le1$⁠, both $x$ and $1-x$ are in $[0,1]$⁠, so

   $$0 \le \alpha \le \frac\pi2, \qquad 0 \le \beta \le \frac\pi2.$$

   The equation says $\beta = \frac\pi2 + 2\alpha$⁠. The right side is at least $\frac\pi2$ (since $\alpha \ge 0$⁠), and the left side is at most $\frac\pi2$⁠. So both must equal $\frac\pi2$⁠:

   $$\alpha = 0 \ \text{ and } \ \beta = \frac\pi2.$$

   $\alpha = \sin^{-1}x = 0$ gives $x = 0$⁠, and then $\beta = \sin^{-1}1 = \frac\pi2$⁠, consistent.

   This argument uses only the principal ranges, so it cannot introduce false roots. It also gives uniqueness for free. Another way to see it: $\sin^{-1}(1-x)$ decreases and $-2\sin^{-1}x$ decreases as $x$ grows, so the left side is strictly decreasing and takes the value $\frac\pi2$ at most once.
   Range of $\sin^{-1}$ on $[0,1]$ is $\left[0,\frac\pi2\right]$ (Basic 1). Lesson: principal values

3. Check x = 0 in the original equation

   $$\sin^{-1}(1-0) - 2\sin^{-1}0 = \frac\pi2 - 0 = \frac\pi2. \ ✓$$
   Even after a clean argument, substituting back is the final confirmation.
   $\sin^{-1}1 = \frac\pi2$⁠, $\sin^{-1}0 = 0$ (Basic 1).

4. The "take sine of both sides" route, and its false root

   Write the equation as $\beta = \frac\pi2 + 2\alpha$ and take the sine of both sides:

   $$\begin{aligned} 1 - x &= \sin\left(\tfrac\pi2 + 2\alpha\right) = \cos2\alpha \\ &= 1 - 2\sin^2\alpha = 1 - 2x^2. \end{aligned}$$

   So $2x^2 - x = 0$⁠, that is $x(2x-1) = 0$⁠, giving $x = 0$ or $x = \frac12$⁠.

   Test $x = \frac12$ in the original equation:

   $$\sin^{-1}\tfrac12 - 2\sin^{-1}\tfrac12 = \tfrac\pi6 - \tfrac\pi3 = -\tfrac\pi6 \ne \tfrac\pi2.$$

   So $x=\frac12$ is not a solution.

   Taking the sine is not reversible. At $x=\frac12$⁠: $\beta = \frac\pi6$⁠, while $\frac\pi2 + 2\alpha = \frac\pi2 + \frac\pi3 = \frac{5\pi}6$⁠. These angles are different, but both have sine $\frac12$⁠, so the equation obtained by taking sines cannot tell them apart. The angle $\frac{5\pi}6$ is outside $\left[-\frac\pi2,\frac\pi2\right]$⁠, the range of $\sin^{-1}$⁠, so it can never equal $\beta$⁠. This is exactly the trap behind option (A).
   Equal sines imply equal angles when both angles are inside $\left[-\frac\pi2,\frac\pi2\right]$ (Basic 2); $\cos2\alpha = 1-2\sin^2\alpha$ (Basic 3).

5. Test every option

   Option	Claim	Verdict	Reason
   (A)	$0, \frac12$	wrong	$0$ works, but $\frac12$ gives $-\frac\pi6$⁠, not $\frac\pi2$ (Step 4).
   (B)	$1, \frac12$	wrong	$x = 1$ gives $\sin^{-1}0 - 2\sin^{-1}1 = -\pi$⁠; and $\frac12$ fails too.
   (C)	$0$	correct	The only solution (Steps 2 and 3).
   (D)	$\frac12$	wrong	$\frac12$ gives $-\frac\pi6$⁠.

   An option listing several values is correct only if every value listed is a solution and no solution is missing. Only (C) passes.

Every fact used, and where it comes from