Let the set of all a ∈ R such that the equation 2x + asinx = 2a - 7...

Given: $\cos 2x + a \cdot \sin x = 2a - 7$

$⇒ 1 - 2\sin^{2}x + a \times \sin x = 2a - 7$

$⇒ a \times \sin x - 2a = - 8 + 2\sin^{2}x$

$⇒ a\left(\sin x - 2\right) = 2\left(\sin^{2}x - 4\right)$

$⇒ a ( \sin x - 2 ) = 2 ( \sin x - 2 ) ( \sin x + 2 )$

$⇒ \sin x = 2\left[\text{Not possible}\right], a = 2 ( \sin x + 2 )$

We know that, $- 1 \le \sin x \le 1$

$⇒ a \in [ 2, 6 ]$

$⇒ p = 2, q = 6$

Now, $r = tan{9}^{°} + cot{9}^{°} - tan{27}^{°} - cot{27}^{°}$

$⇒ r = \dfrac{\cos^{2}{9}^{°} + \sin^{2}{9}^{°}}{sin{9}^{°} \cdot cos{9}^{°}} - \dfrac{\cos^{2}{27}^{°} + \sin^{2}{27}^{°}}{sin{27}^{°} \cdot cos{27}^{°}}$

$⇒ r = \dfrac{1}{sin{9}^{°} \cdot cos{9}^{°}} - \dfrac{1}{sin{27}^{°} \cdot cos{27}^{°}}$

$⇒ r = \dfrac{2}{2 \times sin{9}^{°} \cdot cos{9}^{°}} - \dfrac{2}{2 \times sin{27}^{°} \cdot cos{27}^{°}}$

$⇒ r = \dfrac{2}{sin{18}^{°}} - \dfrac{2}{sin{54}^{°}}$

$⇒ r = 2\left[\dfrac{4}{\sqrt{5} - 1} - \dfrac{4}{\sqrt{5} + 1}\right]$

$⇒ r = 8\left[\dfrac{\sqrt{5} + 1 - \sqrt{5} + 1}{5 - 1}\right]$

$⇒ r = 8\left(\dfrac{2}{4}\right)$

$⇒ r = 4$

$⇒ p \cdot q \cdot r = 2 \times 6 \times 4$

$⇒ p \cdot q \cdot r = 48$