Question

1 - 4 sin 2 x + 6 cos ^2 x = 0

Answer 1

Решение.

Sin2x – 5sinx cosx + 6cos2x = 0

Делим обе части уравнения на cos2x

tg2x – 5tgx + 6 = 0

tgx = t

t2 – 5t + 6=0

t1 = 3 t2 = 2

tgx = 3

tgx = 2

x= arctg3 + πk, k Z

x= arctg2 + πn, n Z

Ответ: x = arctg3 + πk, х = arctg2 + πn, n, k Z