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 $D\ge 0$
$\Rightarrow 16{\cos }^2\left(\alpha -\beta \right)-16\ge 0\Rightarrow 16{\cos }^2\left(\alpha -\beta \right)\ge 16$
$\Rightarrow {\cos }^2\left(\alpha -\beta \right)\ge 1$
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
$4t^2+4t+1=0\Rightarrow {\left(2t+1\right)}^2=0\Rightarrow t=-\frac{1}{2}$
$\Rightarrow \cos \left(\alpha +\beta \right)=-\frac{1}{2}=\cos \frac{2\pi }{3}$
$\Rightarrow \alpha +\beta =\frac{2\pi }{3}$
Now, $\alpha +\beta =\frac{2\pi }{3}$ and $\alpha =\beta \Rightarrow \alpha =\beta =\frac{\pi }{3}$
$\Rightarrow 2\alpha +\beta =\pi$