Решить, : 3 1) 7sin^2x+3sinx*cosx-2cos^2x=4 2) sin2x=2cos^2x

1) 7sin^2(x) + 3sinx*cosx - 2cos^2(x) - 4sin^2(x) - 4cos^2(x) = 0
3sin^2(x) + 3sinx*cosx - 6cos^2(x) = 0
tg^2(x) + tgx - 2 = 0
tgx = -2, x=-arctg(2) + πk
tgx = 1, x=π/4 + πk
2) sin(2x) = 2cos^2(x)
2sinx*cosx - 2cos^2(x) = 0
2cosx*(sinx - cosx) = 0
cosx = 0, x=π/2 + πk
sinx=cosx, tgx = 1, x=π/4 + πk

Думаю ответ таков,но незнаю ли то...