А)log3(2cosx) = t
6t^2-11t+4=0
D = 121 - 4 * 4 * 6 = 121 - 96 = 25
t1 = (11 + 5) / 12 = 1
t2 = (11 - 5) / 12 = 1/2
1)log3(2cosx) = 1
log3(2cosx) = log3(3)
2cosx = 3
cosx = 3/2 (нет решения)
2) log3(2cosx) = 1/2
log3(2cosx) = log3(√3)
2cosx = √3
cosx = √3/2
x = +-Pi/6+2Pi*n
б) 1)-7Pi/2 <= </span>Pi/6+2Pi*n <= -Pi<br> -11Pi/3 <= 2*Pi*n <= -7Pi/6<br> -11/6 <= n <= -7/12<br> n = -1
x = Pi/6 -2*Pi = -11Pi/6
2) -7Pi/2 <= -Pi/6+2Pi*n <= -Pi<br>-10Pi/3 <= 2*Pi*n <= -5Pi/6<br>-10/6 <= n <= -5/12<br>n = -1
x = -Pi/6 -2Pi = -13Pi/6