Question

Решите, если можно подробно

Answer 1

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 <= Pi/6+2Pi*n <= -Pi
     -11Pi/3 <= 2*Pi*n <= -7Pi/6
      -11/6 <= n <= -7/12
 n = -1 
x = Pi/6 -2*Pi = -11Pi/6
2) -7Pi/2 <= -Pi/6+2Pi*n <= -Pi
-10Pi/3 <= 2*Pi*n <= -5Pi/6
-10/6 <= n <= -5/12
n = -1
x = -Pi/6 -2Pi = -13Pi/6