Решите, если можно подробно

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