Step 1: Square the given equation.
Given $ \sin x + a\cos x = b $. Squaring: $ \sin^2 x + a^2 \cos^2 x + 2a\sin x \cos x = b^2 $.
Step 2: Expand (a sin x - cos x)^2.
$ (a\sin x - \cos x)^2 = a^2\sin^2 x + \cos^2 x - 2a\sin x\cos x $.
Step 3: Add the two expanded expressions.
$(\sin x + a\cos x)^2 + (a\sin x - \cos x)^2 $ \[ = \sin^2 x + a^2\cos^2 x + 2a\sin x\cos x + a^2\sin^2 x + \cos^2 x - 2a\sin x\cos x \] The cross terms $ \pm 2a\sin x\cos x $ cancel: \[ = (\sin^2 x + \cos^2 x) + a^2(\cos^2 x + \sin^2 x) = 1 + a^2 \]
Step 4: Substitute and isolate.
Since the left side equals $ b^2 + (a\sin x - \cos x)^2 = 1 + a^2 $, we get: \[ (a\sin x - \cos x)^2 = 1 + a^2 - b^2 = a^2 - b^2 + 1 \]
Step 5: Take the square root.
Taking the non-negative square root: $ |a\sin x - \cos x| = \sqrt{a^2 - b^2 + 1} $.
Step 6: State the final answer.
\[ \boxed{\sqrt{a^2 - b^2 + 1}} \]