Suppose θ ε [0, π 4 ] is a solution of 4 cos θ-3 sin θ=1. Then cos θ is equal to :

$4\left(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\right)-3\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)=1$4(1+tan2θ/21−tan2θ/2​)−3(1+tan22θ​2tan2θ​​)=1 let $\tan \frac{\theta}{2}=\mathrm{t}$tan2θ​=t $\begin{aligned}
& \frac{4-4 t^2-6 t}{1+t^2}=1 \
& 4-4 t^2-6 t=1+t^2
\end{aligned}$$\begin{aligned} & \Rightarrow 5 t^2+6 t-3=0 \ & \Rightarrow t=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)} \ & =\frac{-6 \pm \sqrt{96}}{10} \ & =\frac{-6 \pm 4 \sqrt{6}}{10} \ & t=\frac{-3+2 \sqrt{6}}{5} \ & \cos \theta=\frac{1-t^2}{1+t^2}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^2}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^2}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} \ & =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}} \ & =\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}} \ & =\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}\end{aligned}$