Решите уравнение: 6cos2x-14cos^2 x-7sin2x=0

 

 

 

6cos2x - 14cos²x - 7sin2x = 0 6cos²x - 6sin²x - 14cos²x - 14 sinx·cosx =0 -8cos²x - 6sin²x - 14sinx·cosx = 0 4cos²x + 3sin²x + 7sinx·cosx = 0 :cos²x 4 + 3tg²x + 7tgx = 0 замена tgx = у 3у² + 7у + 4 = 0 D = 49 - 48 = 1 у1 = (-7 -1): 6 = -4/3 у2 = (-7 +1): 6 = -1 tgx1 = -4/3 х1 = -arctg (4/3) + πk k∈Z tgx2 = -1 х1 = -π/4 + πk k∈Z