The value of 2α + β (0 < α, β < (π)/(2)), satisfying the equation...

From the given relation,

$\left(2cos \alpha cos \beta\right)\cos \left(\alpha + \beta\right) = - \dfrac{1}{4}$

\left(\cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right)\right)\cos \left(\alpha + \beta\right) = - \dfrac{1}{4}$<br /><br />Let,~$\cos \left(\alpha + \beta\right) = t$<br /><br />\left(t + \cos \left(\alpha - \beta\right)\right)t = - \dfrac{1}{4}$

$⇒ 4t^{2} + 4tcos\left(\alpha - \beta\right) + 1 = 0$

The roots will be real if $D \ge 0$

$⇒ 16cos^{2}\left(\alpha - \beta\right) - 16 \ge 0 ⇒ 16cos^{2}\left(\alpha - \beta\right) \ge 16$

$⇒ cos^{2}\left(\alpha - \beta\right) \ge 1 ⇒ cos^{2}\left(\alpha - \beta\right) = 1 ⇒ \alpha - \beta = 0 ⇒ \alpha = \beta$

Then the above quadratic equation becomes

$4t^{2} + 4t + 1 = 0 ⇒ {\left(2t + 1\right)}^{2} = 0 ⇒ t = - \dfrac{1}{2}$

$⇒ \cos \left(\alpha + \beta\right) = - \dfrac{1}{2} = cos\dfrac{2\pi}{3} ⇒ \alpha + \beta = \dfrac{2\pi}{3}$

Now, $\alpha + \beta = \dfrac{2\pi}{3}$ and $\alpha = \beta ⇒ \alpha = \beta = \dfrac{π}{3}$

$⇒ 2\alpha + \beta = \pi$