Question

If $2cos\Theta +sin\Theta =1,thencos\Theta +6sin\Theta =?$

Answer 1

We have,

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

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

Squaring both side and we get,

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

$\Rightarrow 4{\cos }^2\theta =1^2+{\sin }^2\theta -2\sin \theta$

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

$\Rightarrow 4-4{\sin }^2\theta =1+{\sin }^2\theta -2\sin \theta$

$\Rightarrow -4{\sin }^2\theta -{\sin }^2\theta +2\sin \theta +4-1=0$

$\Rightarrow -5{\sin }^2\theta +2\sin \theta +3=0$

$\Rightarrow 5{\sin }^2\theta -2\sin \theta -3=0$

$\Rightarrow 5{\sin }^2\theta -(5-3)\sin \theta -3=0$

$\Rightarrow 5{\sin }^2\theta -5\sin \theta +3\sin \theta -3=0$

$\Rightarrow 5\sin \theta (\sin \theta -1)+3(\sin \theta -1)=0$

$\Rightarrow (\sin \theta -1)(5\sin \theta +3)=0$

$\Rightarrow \sin \theta -1=0,5\sin \theta =-3$

$\Rightarrow \sin \theta =1,\sin \theta =-\frac{3}{5}$

$\Rightarrow \sin \theta =\sin \frac{\pi }{2},\sin \theta =-\frac{3}{5}$

$\Rightarrow \theta =\frac{\pi }{2}$

Put the value of $\theta$ in equation (1) and we get,

$\cos \theta +6\sin \theta =?$

$=\cos \frac{\pi }{2}+6\sin \frac{\pi }{2}$

$=0+6\times 1$

$=6$

Hence, this is the answer.