36^sinx+36^cos(x+п/2)=37/6

Решите задачу:

$36^{sinx}+36^{cos(x+ \frac{\pi}{2} )}= \frac{37}{6} \\36^{sinx}+36^{-sinx}=\frac{37}{6} \\36^{sinx}=y,\ y \in (0;+\infty) \\y+ \frac{1}{y} =\frac{37}{6} \\y^2-\frac{37}{6} *y+1=0 \\6y^2-37y+6=0 \\D=1369-144=1225=35^2 \\y_1= \frac{37+35}{12} =6 \\y_2= \frac{37-35}{12} = \frac{1}{6} =6^{-1} \\36^{sinx}=6 \\6^{2sinx}=6^1 \\2sinx=1 \\sinx= \frac{1}{2} \\x_1= \frac{\pi}{6} +2\pi n,\ n \in Z \\x_2= \frac{5\pi}{6} +2\pi n,\ n \in Z \\36^{sinx}=6^{-1} \\6^{2sinx}=6^{-1} \\2sinx=-1$
$sinx=- \frac{1}{2} \\x_3= -\frac{\pi}{6} +2\pi n,\ n \in Z \\x_4= -\frac{5\pi}{6} +2\pi n,\ n \in Z$