Mathematics - Trigonometric Equation Question with Solution | TestHub

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