Given $2 \cos \theta+\sin \theta=1$

Squaring both sides, we get $(2 \cos \theta+\sin \theta)^{2}=1^{2}$

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

$\Rightarrow 3 \cos ^{2} \theta+\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+4 \sin \theta \cos \theta$ $=1$

$\Rightarrow 3 \cos ^{2} \theta+1+4 \sin \theta \cos \theta=1$

$\Rightarrow 3 \cos ^{2} \theta+4 \sin \theta \cos \theta=0$

$\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0$

$\Rightarrow 3 \cos \theta+4 \sin \theta=0$

$\Rightarrow 3 \cos \theta=-4 \sin \theta$

$\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^{2} \theta-1}=\frac{-3}{4}$

$(\because \tan \theta=\sqrt{\sec ^{2} \theta-1})$

$\Rightarrow \sec ^{2} \theta-1=$ $\left(\frac{-3}{4}\right)^{2}=\frac{9}{16}$

$\Rightarrow \sec ^{2} \theta=\frac{9}{16}+1=$ $\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4}$

or ${\cos \theta=\frac{4}{5}}$    ……. $(1)$

Now, $\sin ^{2} \theta+\cos ^{2} \theta=1$

$\Rightarrow \sin ^{2} \theta+\left(\frac{4}{5}\right)^{2}=1$

$\sin ^{2} \theta+\frac{4}{5}=1 \Rightarrow \sin ^{2} \theta=1-\frac{16}{25}=\frac{9}{25}$

$\sin \theta=\pm \frac{3}{5}$     …….. $(2)$

Taking $\quad\left(\sin \theta=-\frac{3}{5}\right) \quad$ because $\left(\sin \theta=\frac{3}{5}\right)$ cannot satisfy the given equation.

Therefore; $7 \cos \theta+6 \sin \theta$

$=7 \times \frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}-\frac{18}{5}=2$