Question

If the equation $2^{2 \sin ^2 x+8}-p \cdot 2^{(\sin x+2)^2}+2^{8 \sin x}=0$ possesses a solution, then find the number of integral values of p.

Answer 1

$2^{2 \sin ^2 x +8}- p \cdot 2^{(\sin x +2)^2}+2^{8 \sin x }=0$
On divide the equation by $2^{8 \sin x}$, we get
$2^{2 \sin ^2 x-8 \sin x+8}-p \cdot 2^{\sin ^2 x-4 \sin x+4}+1=0 $
$2^{2\left(\sin ^2 x-4 \sin x+4\right)}+1=p \cdot 2^{\left(\sin ^2 x-4 \sin x+4\right)}$
Let
$t =2^{\sin ^2 x -4 \sin x +4}=2^{(\sin x -2)^2} $
$t \in\left[2,2^9\right] $
$t ^2+1= pt$
$p = t +\frac{1}{ t }$
$\therefore p \in\left[2+\frac{1}{2}, 2^9+\frac{1}{2^9}\right] \Rightarrow$ Integral values of ' $p$ ' are $3,4,5$, 512
$\therefore$ Number of integral values of $p$ are $512-2=510$.