To solve the equation \( \sin x + a \cos x = b \), we can proceed with the following steps: 1. Square Both Sides: \[ (\sin x + a \cos x)^2 = b^2 \] Expand the left side: \[ \sin^2 x + a^2 \cos^2 x + 2a \sin x \cos x = b^2 \] 2. Use Trigonometric Identity: Replace \( \sin^2 x \) with \( 1 - \cos^2 x \): \[ (1 - \cos^2 x) + a^2 \cos^2 x + 2a \sin x \cos x = b^2 \] Simplify: \[ (a^2 - 1) \cos^2 x + 2a \sin x \cos x = b^2 - 1 \] 3. Express in Terms of \( \cos x \): \[ 2a \sin x \cos x = b^2 - 1 + (1 - a^2) \cos^2 x \quad \text{(1)} \] 4. Define \( k \): Let \( k = |a \sin x - \cos x| \). Then: \[ k^2 = (a \sin x - \cos x)^2 \] Expand the right side: \[ k^2 = a^2 \sin^2 x + \cos^2 x - 2a \sin x \cos x \] Replace \( \sin^2 x \) with \( 1 - \cos^2 x \): \[ k^2 = a^2 (1 - \cos^2 x) + \cos^2 x - 2a \sin x \cos x \] Simplify: \[ k^2 = a^2 + (1 - a^2) \cos^2 x - 2a \sin x \cos x \] 5. Substitute from Equation (1): From equation (1), we have: \[ 2a \sin x \cos x = b^2 - 1 + (1 - a^2) \cos^2 x \] Substitute into the expression for \( k^2 \): \[ k^2 = a^2 + (1 - a^2) \cos^2 x - [b^2 - 1 + (1 - a^2) \cos^2 x] \] Simplify: \[ k^2 = a^2 - b^2 + 1 \] Therefore, the value of \( k\) is: \[ \boxed{\sqrt{a^2 - b^2 + 1}} \]