اوجد مجموعة الحل

2^جاس +3^جاس +6^جاس=1

$\large {\color{Red} 2^{sinx}+3^{sinx}+b^{sinx}=1}$
${\color{DarkBlue} -1\leq sinx\leq 1 \rightarrow \frac{1}{2}\leq 2^{sinx}\leq 2 ;;\frac{1}{3}\leq 3^{sinx}\leq 3;;\frac{1}{b}\leq b^{sinx}\leq b}$
${\color{DarkBlue}\rightarrow \frac{1}{2}+\frac{1}{3}+\frac{1}{b}\leq2^{sinx}+3^ {sinx}+b^{sinx}\leq 11 }$
${\color{DarkBlue} \Rightarrow 1\leq 2^{sinx}+3^{sinx}+b^{sinx}\leq 11}$
${\color{DarkBlue} \Rightarrow 2^{sinx}+3^{sinx}+b^{sinx}=1 if;sinx=-1\rightarrow x=-(\frac{\pi}{2})+(2\pi)k}$