Question

If $2\cos x+\sin x=1,$ then the sum of the values of $7\cos x+6\sin x$ is

Answer 1

Using $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}},\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$
$\therefore \sin x+2\cos x=1$
$\Rightarrow \left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+2\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)=1$
$\Rightarrow 3{\tan }^2\frac{x}{2}-2\tan \frac{x}{2}-1=0$
$\Rightarrow (3\tan \frac{x}{2}+1)(\tan \frac{x}{2}-1)=0$
$\Rightarrow \tan \frac{x}{2}=1$ or $-\frac{1}{3}$
Now $7\cos x+6\sin x$
$=7\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+6\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)$
When $\tan \frac{x}{2}=1\Rightarrow 7\cos x+6\sin x=6$
And when $\tan \frac{x}{2}=-\frac{1}{3}\Rightarrow 7\cos x+6\sin x=2$
Therefore sum of value of solution is $6+2=8$