Question

If $(2\cos x+\sin x)=1$, then sum of all possible value of $(7\cos x+6\sin x)$ is__________

Answer 1

Given $2\cos x+\sin x=1⟹\frac{2(1-{\tan }^2\frac{x}{2})}{1+{\tan }^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=1$

$⟹2-2{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}=1+{\tan }^2\frac{x}{2}$

$⟹3{\tan }^2\frac{x}{2}-2\tan \frac{x}{2}-1=0$

$⟹3{\tan }^2\frac{x}{2}-3\tan \frac{x}{2}+\tan \frac{x}{2}-1=0$

$⟹(\tan \frac{x}{2}-1)(3\tan \frac{x}{2}+1)=0$

So $\tan \frac{x}{2}=1,-\frac{1}{3}$

When $\tan \frac{x}{2}=1$

then $7\cos x+6\sin x=7(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}})+6(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}})=7\times \frac{1-1}{1+1}+6\times \frac{2\times 1}{1+1}=0+6=6$

When $\tan \frac{x}{2}=-\frac{1}{3}$

then $7\cos x+6\sin x=7(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}})+6(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}})=7\times \frac{1-\frac{1}{9}}{1+\frac{1}{9}}+6\times \frac{-2\times \frac{1}{3}}{1+\frac{1}{9}}=7\times \frac{8}{10}-6\times \frac{6}{10}=2$

The sum of all possible values is $6+2=8$