If an angle A of a ABC satisfies 5cosA + 3 = 0 , then the roots of...

We have,

$5cosA + 3 = 0 ⇒ cosA = - \dfrac{3}{5}$

Clearly, $A \in \left(90 ° , 180 °\right)$

Now, roots of equation

$9{x}^{2} + 27x + 20 = 0$

$⇒ x = \dfrac{- 27 \pm \sqrt{729 - 720}}{18} = \dfrac{- 27 \pm 3}{18}$

$⇒ x = - \dfrac{5}{3}, - \dfrac{4}{3}$

Then, $- \dfrac{5}{3} = secA$ and $- \dfrac{4}{3} = tanA$.

Hence, the roots are $secA$ and $tanA$.