Given $2 \cos \theta+\sin \theta=1$
Squaring both sides, we get
$$
\begin{aligned}
&(2 \cos \theta+\sin \theta)^2=1^2 \\
&\Rightarrow 4 \cos ^2 \theta+\sin ^2 \theta+4 \sin \theta \cos \theta=1 \\
&\Rightarrow 3 \cos ^2 \theta+\left(\cos ^2 \theta+\sin ^2 \theta\right)+4 \sin \theta \cos \theta \\
&=1 \\
&\Rightarrow 3 \cos ^2 \theta+\not \lambda+4 \sin \theta \cos \theta=1 \\
&\Rightarrow 3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\
&\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0 \\
&\Rightarrow 3 \cos \theta+4 \sin \theta=0 \\
&\Rightarrow 3 \cos \theta=-4 \sin \theta \\
&\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^2 \theta-1}=\frac{-3}{4} \\
&\left(\because \tan ^2 \theta=\sqrt{\sec ^2 \theta-1}\right) \\
&\Rightarrow \sec ^2 \theta-1=\left(\frac{-3}{4}\right)^2=\frac{9}{16} \\
&\Rightarrow \sec 2 \theta=\frac{9}{16}+1=\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4}
\end{aligned}
$$
or $\cos \theta=\frac{4}{5}$
Now, $\sin ^2 \theta+\cos ^2 \theta=1$
$$
\begin{aligned}
&\Rightarrow \sin ^2 \theta+\left(\frac{4}{5}\right)^2=1 \\
&\sin ^2 \theta+\frac{4}{5}=1 \Rightarrow \sin ^2 \theta=1-\frac{16}{25}=\frac{9}{25} \\
&\sin \theta=\pm \frac{3}{5}
\end{aligned}
$$
Taking $\quad\left(\sin \theta=+\frac{3}{5}\right) \quad$ because $\left(\sin \theta=-\frac{3}{5}\right)$ cannot satisfy the given equation.
Therefore; $7 \cos \theta+6 \sin \theta$
$$
=7 \times \frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}
$$