Question

The value of $2\alpha +\beta \left(0 < \alpha , \beta < \frac{\pi }{2}\right)$ , satisfying the equation $\cos\alpha \cos\beta \cos\left(\alpha + \beta \right)=-\frac{1}{8}$ is equal to

• A. $\frac{5}{6}\pi $
• B. $\frac{\pi }{2}$
• C. $\pi $
• D. $\frac{7 \pi }{12}$

Answer 1

Answer: (C)

From the given relation, $\left(2 \cos \alpha \cos \beta \right)\cos\left(\alpha + \beta \right)=-\frac{1}{4}$
$\left(\cos \left(\alpha + \beta \right) + \cos \left(\alpha - \beta \right)\cos\left(\alpha + \beta \right)=-\frac{1}{4}\right)$
Let, $\cos\left(\alpha + \beta \right)=t$
$\left(t + \cos \left(\alpha - \beta \right)t=-\frac{1}{4}\right)$
$\Rightarrow 4t^{2}+4t\cos\left(\alpha - \beta \right)+1=0$
The roots will be real if $D\geq 0$
$\Rightarrow 16\cos^{2}\left(\alpha - \beta \right)-16\geq 0\Rightarrow 16\cos^{2}\left(\alpha - \beta \right)\geq 16$
$\Rightarrow \cos^{2}\left(\alpha - \beta \right)\geq 1$
Only possible when
$\Rightarrow \left(\cos\right)^{2}\left(\alpha - \beta \right)=1\Rightarrow \alpha -\beta =0\Rightarrow \alpha =\beta $
Then the above quadratic equation becomes
$4t^{2}+4t+1=0\Rightarrow \left(2 t + 1\right)^{2}=0\Rightarrow t=-\frac{1}{2}$
$\Rightarrow \cos\left(\alpha + \beta \right)=-\frac{1}{2}=\cos\frac{2 \pi }{3}$
$\Rightarrow \alpha +\beta =\frac{2 \pi }{3}$
Now, $\alpha +\beta =\frac{2 \pi }{3}$ and $\alpha =\beta \Rightarrow \alpha =\beta =\frac{\pi }{3}$
$\Rightarrow 2\alpha +\beta =\pi $