Решить уравнение

7cos3x+cos6x-3=0

6sin^2x+5cosx-7=0

7cos3x + 2cos^2 (3x) - 1 - 3 = 0

7cos3x + 2cos^2 (3x) - 4 = 0

cos3x = t

7t + 2 t^2 - 4=0

2t^2 + 7t - 4 = 0

D = 81

cos3x = 0.5 и cos3x=-4, корней нет

3x = + - (п/3) + 2 пN

х = + - (п/9) + (2/3) пN

6 (1-cos^2 (x)) + 5cosx - 7=0

6 - 6cos^2 (x) + 5 cosx - 7 = 0

6cos^2 (x) - 5cosx + 1 = 0

cosx=t

6t^2 - 5t + 1=0

D = 1

cosx=0.5

x = + - (п/3) + 2 пN