А) По свойствам логарифма
log3 (sin^2 x) = 2*log3 (sin x)
Сделаем замену t = log3 (sin x)
t^2 + 2t = log3(2)*t
t^2 + t*(2 - log3(2) ) = 0
t*(t + 2 - log3(2) ) = 0
1) t = log3 (sin x) = 0
sin x = 1
x1 = pi/2 + 2pi*n
2) t = log3(2) - 2
log3 (sin x) = log3(2) - log3(9) = log3(2/9)
sin x = 2/9
x2 = arcsin(2/9) + 2pi*k
x3 = pi - arcsin(2/9) + 2pi*k
Б) arcsin(2/9)≈2/9=0,22 < pi/3, поэтому в [pi/3; 2pi] попадают корни:
x1 = pi/2; x2 = pi - arcsin(2/9)