Without using tables, find: $\;\;\;$ $\cot \left(\dfrac{\alpha}{2}\right)$, $\;$ given $\;$ $20 \sin^2 \alpha + 21 \cos \alpha - 24 = 0$, $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$
Given $\;$ $20 \sin^2 \alpha + 21 \cos \alpha - 24 = 0$
i.e. $\;$ $20 \left(1 - \cos^2 \alpha\right) + 21 \cos \alpha - 24 = 0$
i.e. $\;$ $- 20 \cos^2 \alpha + 21 \cos \alpha - 4 = 0$
i.e. $\;$ $20 \cos^2 \alpha - 21 \cos \alpha + 4 = 0$
i.e. $\;$ $20 \cos^2 \alpha - 16 \cos \alpha - 5 \cos \alpha + 4 = 0$
i.e. $\;$ $5 \cos \alpha \left(4 \cos \alpha - 1\right) - 4 \left(4 \cos \alpha - 1\right) = 0$
i.e. $\;$ $\left(5 \cos \alpha - 4\right) \left(4 \cos \alpha - 1\right) = 0$
i.e. $\;$ $5 \cos \alpha - 4 = 0$ $\;$ or $\;$ $4 \cos \alpha - 1 = 0$
i.e. $\;$ $\cos \alpha = \dfrac{4}{5}$ $\;$ or $\;$ $\cos \alpha = \dfrac{1}{4}$
Since $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$
$\implies$ $\alpha$ is in the fourth quadrant.
$\implies$ $\cos \alpha$ is positive and $\sin \alpha$ is negative.
Now, when $\;$ $\cos \alpha = \dfrac{4}{5}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{16}{25}} = \pm \dfrac{3}{5}$
When $\;$ $\cos \alpha = \dfrac{1}{4}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{1}{16}} = \pm \dfrac{\sqrt{15}}{4}$
$\because \;$ $\alpha$ is in the fourth quadrant, $\;$ $\sin \alpha = \dfrac{-3}{5}$ $\;$ or $\;$ $\sin \alpha = \dfrac{- \sqrt{15}}{4}$
Further, since $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$
$\implies$ $\dfrac{7 \pi}{8} < \dfrac{\alpha}{2} < \pi$
$\implies$ $\dfrac{\alpha}{2}$ lies in the second quadrant.
$\implies$ $\sin \left(\dfrac{\alpha}{2}\right)$ is positive and $\cos \left(\dfrac{\alpha}{2}\right)$ is negative.
Now, $\;$ $\cos \alpha = 2 \cos^2 \left(\dfrac{\alpha}{2}\right) - 1$
$\implies$ $\cos^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 + \cos \alpha}{2} = \dfrac{1 + \dfrac{4}{5}}{2} = \dfrac{9}{10}$ $\;$ or $\;$ $\cos^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 + \dfrac{1}{4}}{2} = \dfrac{5}{8}$
$\implies$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm 3}{\sqrt{10}}$ $\;$ or $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm \sqrt{5}}{2 \sqrt{2}}$
$\because \;$ $\dfrac{\alpha}{2}$ is in the second quadrant, $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$ $\;$ or $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{- \sqrt{5}}{2 \sqrt{2}}$
Now, $\;$ $\sin \alpha = 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right)$ $\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\sin \alpha}{2 \cos \left(\dfrac{\alpha}{2}\right)}$
When $\;$ $\sin \alpha = \dfrac{-3}{5}$, $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-3}{5}}{-2 \times \dfrac{3}{\sqrt{10}}} = \dfrac{1}{\sqrt{10}}$
When $\;$ $\sin \alpha = \dfrac{-\sqrt{15}}{4}$, $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-5}{2 \sqrt{2}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-\sqrt{15}}{4}}{-2 \times \dfrac{\sqrt{5}}{2\sqrt{2}}} = \dfrac{3}{2\sqrt{6}}$
Now, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\cos \left(\dfrac{\alpha}{2}\right)}{\sin \left(\dfrac{\alpha}{2}\right)}$
When $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{1}{\sqrt{10}}$, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-3}{\sqrt{10}}}{\dfrac{1}{\sqrt{10}}} = -3$
When $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-\sqrt{5}}{2\sqrt{2}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{3}{2\sqrt{6}}$, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{- \sqrt{5}}{2 \sqrt{2}}}{\dfrac{3}{2 \sqrt{6}}} = \dfrac{-\sqrt{15}}{3}$