Given that: \[ \frac{3}{5} \sin x + \frac{4}{5} \cos x = 1 \] Since \( \sin^2 x + \cos^2 x = 1 \), so: \[ \sin x = \frac{3}{4}, \quad \cos x = \frac{4}{5} \] \[ \Rightarrow \frac{1 - \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} = \frac{4}{5} \] or: \[ \tan \frac{x}{2} = \frac{1}{3} \] \[ \Rightarrow 6 \tan \frac{x}{2} - 9 \tan^2 \frac{x}{2} = \frac{6}{3} - \frac{9}{9} = 1 \]