6×cos2x× sinx + 7× sin2x = 0
6cos2x × sinx + 14sinx × cosx = 0
2sinx × ( 3cos2x + 7cosx ) = 0
_____________________
sinx = 0
x = πn, n € Z
_____________________
3cos2x + 7cosx = 0
3( 2cos²x - 1 ) + 7cosx = 0
6cos²x + 7cosx - 3 = 0
Сделаем замену:
Пусть cosx = t , t € [ -1; 1 ]
6t² + 7t - 3 = 0
D = 7² - 4×6×(-3) = 49 + 72 = 121 = 11²
t1 = -3/2 = -1,5 - не подходит по условию
t2 = 1/3
cosx = 1/3
x = arccos(1/3) + 2πk, k € Z
x = - arccos(1/3) + 2πm, m € Z
Ответ: πn, n € Z; +- arccos(1/3) + 2πk, k € Z