Sin2xtgx-tgx+sin2x=1

Question

Sin2xtgx-tgx+sin2x=1

Answer 1

Sin2xtgx - tgx + sin2x = 1
sin2xtgx - tgx + sin2x - 1 = 0
tgx(sin2x - 1) + (sin2x - 1)
(tgx + 1)(sin2x - 1) = 0
tgx + 1 = 0
tgx = -1
x = -π/4 + πn, n ∈ Z.
sin2x - 1 = 0
sin2x = 1
2x = π/2 + 2πn, n ∈ Z
x = π/4 + πn, n ∈ Z

Comments

И снова спасибо :)