Question

If $\alpha$ is a root of equation $(2\sin x-\cos x)(1+\cos x)={\sin }^{2}x$, $\beta$ is a root of equation $3{\cos }^{2}x-10\cos x+3=0$ and $\gamma$ is a root of equation $1-\sin 2x=\cos x-\sin x;0\le \alpha ,\beta ,\gamma \le \frac{\pi }{2},$ then $\cos \alpha +\cos \beta +\cos \gamma$ can be equal to

• A. $\frac{3\sqrt{6}+2\sqrt{2}+6}{6\sqrt{2}}$
• B. $\frac{3\sqrt{3}-8}{6}$
• C. $\frac{3\sqrt{3}+2}{6}$
• D. None of these

Answer 1

The correct option is A $\frac{3\sqrt{6}+2\sqrt{2}+6}{6\sqrt{2}}$

$(2\sin x-\cos x)(1+\cos x)={\sin }^{2}x\Rightarrow (2\sin x-\cos x)(1+\cos x)=1-{\cos }^{2}x\Rightarrow (1+\cos x)(2\sin x-\cos x-1+\cos x)=0\Rightarrow (1+\cos x)(2\sin x-1)=0$
$\Rightarrow \cos x=-1$ or $\sin x=\frac{1}{2}$
Since, $0\le \alpha \le \frac{\pi }{2}$
$\Rightarrow \sin \alpha =\frac{1}{2},\cos \alpha =\frac{\sqrt{3}}{2}...\left(1\right)$

$3{\cos }^{2}x-10\cos x+3=0\Rightarrow (3\cos x-1)(\cos x-3)=0\Rightarrow \cos x=\frac{1}{3}(\because \cos x\in [-1,1\left]\right)\therefore \cos \beta =\frac{1}{3},\sin \beta =\frac{2\sqrt{2}}{3}...\left(2\right)$

$1-\sin 2x=\cos x-\sin x$
$\Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=\cos x-\sin x$
$\Rightarrow (\cos x-\sin x{)}^2=\cos x-\sin x$
$\Rightarrow (\cos x-\sin x)(\cos x-\sin x-1)=0$
$\Rightarrow \sin x=\cos x\text{or}\cos x-\sin x=1$
$\Rightarrow \sin \gamma =\cos \gamma =\frac{1}{\sqrt{2}}...\left(3\right)$
Or,
$\cos x-\sin x=1\Rightarrow \cos x=1+\sin x$
$\Rightarrow \cos x=1,\sin x=0(\because 0\le \gamma \le \frac{\pi }{2})$
$\Rightarrow \cos \gamma =1,\sin \gamma =0...\left(4\right)$

$\therefore \cos \alpha +\cos \beta +\cos \gamma$
$=\frac{\sqrt{3}}{2}+\frac{1}{3}+\frac{1}{\sqrt{2}}\text{(or)}\frac{\sqrt{3}}{2}+\frac{1}{3}+1$
$=\frac{3\sqrt{6}+2\sqrt{2}+6}{6\sqrt{2}}\text{(or)}\frac{3\sqrt{3}+8}{6}$