4sin2x+10cos^2x=1 помогите решить

4sin2x + 10cos²x = 18sinxcosx + 10cos²x = sin²x + cos²xsin²x - 8sinxcosx - 9cos²x = 0    |:cos²xtg²x - 8tgx - 9 = 0 tg²x - 8tgx + 16 - 25 = 0(tgx - 4)² - 5² = 0(tgx - 4 - 1)(tgx - 4 + 5) = 0(tgx - 5)(tgx + 1) = 0tgx = 5 или tgx = -1x = arctg5 + πn, n ∈ Z x = -π/4 + πn, n ∈ ZОтвет: x = arctg5 + πn, n ∈ Z; x = -π/4 + πn, n ∈ Z.