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 $\sin (\alpha -\beta )$ is equal to

• A. $\frac{1-\sqrt{6}}{6}$
• B. $\frac{1-2\sqrt{3}}{6}$
• C. $\frac{1-2\sqrt{6}}{6}$
• D. $\frac{\sqrt{3}-2\sqrt{2}}{6}$

Answer 1

The correct option is C $\frac{1-2\sqrt{6}}{6}$

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

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

Now,
$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta =\frac{1}{2}\times \frac{1}{3}-\frac{\sqrt{3}}{2}\times \frac{2\sqrt{2}}{3}=\frac{1-2\sqrt{6}}{6}$