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AI-তৈরি নমুনা উত্তর

Given equation:
$\cos^{-1}x - \sin^{-1}x = \sin^{-1}(1 - x)$

We know that $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$.
Substituting this into the equation:
$\frac{\pi}{2} - 2\sin^{-1}x = \sin^{-1}(1 - x)$

Taking the sine of both sides:
$\sin\left(\frac{\pi}{2} - 2\sin^{-1}x\right) = 1 - x$
$\cos(2\sin^{-1}x) = 1 - x$

Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$ where $\theta = \sin^{-1}x$:
$1 - 2x^2 = 1 - x$
$2x^2 - x = 0$
$x(2x - 1) = 0$

Thus, $x = 0$ or $x = \frac{1}{2}$.

Checking solutions:

• For $x = 0$:
  LHS = $\cos^{-1}(0) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
  RHS = $\sin^{-1}(1 - 0) = \sin^{-1}(1) = \frac{\pi}{2}$
  (LHS = RHS)

• For $x = \frac{1}{2}$:
  LHS = $\cos^{-1}\left(\frac{1}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
  RHS = $\sin^{-1}\left(1 - \frac{1}{2}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
  (LHS = RHS)

Therefore, the solutions are $x = 0, \frac{1}{2}$.