Решите уравнение:

3 + 1 = 7sinx

cos^2x [cosx]

3/cos²x + 1 = 7sinx / (cosx) |•cos²x

3+cos²x=7sinxcosx

3sin²x-7sinxcosx+4cos²x=0 |:cos²x

3tg²x-7tgx+4=0

D=49-48=1

tgx = (7+1) / 6=4/3

x=arctg4/3+πn, n € Z

tgx = (7-1) / 6=1

x=π/4+πn, n € Z