$3\sin x+4\cos x=5 \Rightarrow 6 \tan \frac{x}{2}- 9 {\tan }^2\frac{x}{2}=$

$\begin{array}{rcl}3\sin \mathrm{x}+4\cos \mathrm{x}& =& 5\\ 3\left(\frac{2\tan {\displaystyle \frac{\mathrm{x}}{2}}}{1+{\tan }^2 {\displaystyle \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)+4\left(\frac{1-{\tan }^2 {\displaystyle \frac{\mathrm{x}}{2}}}{1+{\tan }^2 {\displaystyle \frac{\mathrm{x}}{2}}}\right)& =& 5\\ 6\tan \frac{\mathrm{x}}{2}+4-4{\tan }^2 \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& =& 5+5\tan \frac{\mathrm{x}}{2}\\ 6\tan \frac{\mathrm{x}}{2}-9{\tan }^2 \frac{\mathrm{x}}{2}& =& 1\end{array}$