If $2\cos \theta +\sin \theta =1,(\theta \ne (4k+1)\pi /2$$k\in \mathbf{I}\text{ ) }$then $7\cos \theta +6\sin \theta$ is equal to

$2\cos \theta +\sin \theta =1\Rightarrow 4{\cos }^2\theta =(1-\sin \theta {)}^2$

$\begin{array}{l}\Rightarrow 4\left(1-{\sin }^2\theta \right)=(1-\sin \theta {)}^2\\ \Rightarrow 4(1+\sin \theta )=1-\sin \theta \\ \Rightarrow \sin \theta =-3/5\end{array}$

As $2\cos \theta +\sin \theta =1\text{, }$we get $\cos \theta =4/5\text{. }$

Thus $7\cos \theta +6\sin \theta =7\left(\frac{4}{5}\right)+6\left(\frac{-3}{5}\right)=2$