Question

If $2\cos \theta +\sin \theta =1(\theta \ne \frac{\pi }{2})$, then $7\cos \theta +6\sin \theta$ is equal to:

• A. $\frac{11}{2}$
• B. $\frac{46}{5}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is D $2$

$2\cos \theta +\sin \theta =1$ $\theta \ne \frac{\pi }{2}$

$2\cos \theta =1-\sin \theta$

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

Let $\sin \theta =x$

$4-4x^2=1+x^2-2x$

$5x^2-2x-3=0$

$5x^2-5x+3x-3=0$

$(x-1)(5x+3)=0$

$x=1\Rightarrow \sin \theta =1\Rightarrow \theta =\frac{\pi }{2}$ but given $\theta \ne \frac{\pi }{2}$

$\therefore x=\frac{-3}{5}$

$\sin \theta =\frac{-3}{5}$

$\therefore \cos \theta =\frac{4}{5}$

$\therefore 7\cos \theta +6\sin \theta =7\times \frac{4}{5}+6\times \frac{-3}{5}=\frac{10}{5}=2$