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

Answer: 510

$2^{2{\sin }^{2}x+8}-p\cdot 2^{(\sin x+2{)}^2}+2^{8\sin x}=0$
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 $512-2=510$.