If 2 cos θ+sin θ=1 (θ ≠ π 2 ), then 7 cos θ+6 sin θ is equal to:

Given $2 \cos \theta+\sin \theta=1$2cosθ+sinθ=1 Squaring both sides, we get $$\begin{aligned} &(2 \cos \theta+\sin \theta)^2=1^2 \\ &\Rightarrow 4 \cos ^2 \theta+\sin ^2 \theta+4 \sin \theta \cos \theta=1 \\ &\Rightarrow 3 \cos ^2 \theta+\left(\cos ^2 \theta+\sin ^2 \theta\right)+4 \sin \theta \cos \theta \\ &=1 \\ &\Rightarrow 3 \cos ^2 \theta+\not \lambda+4 \sin \theta \cos \theta=1 \\ &\Rightarrow 3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ &\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0 \\ &\Rightarrow 3 \cos \theta+4 \sin \theta=0 \\ &\Rightarrow 3 \cos \theta=-4 \sin \theta \\ &\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^2 \theta-1}=\frac{-3}{4} \\ &\left(\because \tan ^2 \theta=\sqrt{\sec ^2 \theta-1}\right) \\ &\Rightarrow \sec ^2 \theta-1=\left(\frac{-3}{4}\right)^2=\frac{9}{16} \\ &\Rightarrow \sec 2 \theta=\frac{9}{16}+1=\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4} \end{aligned}$$​(2cosθ+sinθ)2=12⇒4cos2θ+sin2θ+4sinθcosθ=1⇒3cos2θ+(cos2θ+sin2θ)+4sinθcosθ=1⇒3cos2θ+λ+4sinθcosθ=1⇒3cos2θ+4sinθcosθ=0⇒cosθ(3cosθ+4sinθ)=0⇒3cosθ+4sinθ=0⇒3cosθ=−4sinθ⇒4−3​=tanθ=sec2θ−1​=4−3​(∵tan2θ=sec2θ−1​)⇒sec2θ−1=(4−3​)2=169​⇒sec2θ=169​+1=1625​⇒secθ=45​​ or $\cos \theta=\frac{4}{5}$cosθ=54​ Now, $\sin ^2 \theta+\cos ^2 \theta=1$sin2θ+cos2θ=1 $$\begin{aligned} &\Rightarrow \sin ^2 \theta+\left(\frac{4}{5}\right)^2=1 \\ &\sin ^2 \theta+\frac{4}{5}=1 \Rightarrow \sin ^2 \theta=1-\frac{16}{25}=\frac{9}{25} \\ &\sin \theta=\pm \frac{3}{5} \end{aligned}$$​⇒sin2θ+(54​)2=1sin2θ+54​=1⇒sin2θ=1−2516​=259​sinθ=±53​​ Taking $\quad\left(\sin \theta=+\frac{3}{5}\right) \quad$(sinθ=+53​) because $\left(\sin \theta=-\frac{3}{5}\right)$(sinθ=−53​) cannot satisfy the given equation. Therefore; $7 \cos \theta+6 \sin \theta$7cosθ+6sinθ $$=7 \times \frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}$$=7×54​+6×53​=528​+518​=546​