Ответы на вопрос:
1+ 2sin2x + 2cos²x = 0 sin²x + cos²x + 4sinxcosx + 2cos²x = 0 sin²x + 4sinxcosx + 3cos²x = 0 |: cos²x tg²x + 4tx + 3 = 0 tg²x + 4tgx + 4 - 1 = 0 (tgx + 2)² - 1² = 0 (tgx + 2 - 1)(tgx + 2 + 1) = 0 (tgx - 1)(tgx + 3) = 0 1) tgx - 1 = 0 tgx = 1 x = π/4 + πn, n ∈ z 2) tgx + 3 = 0 tgx = -3 x = arctg(-3) + πk, k ∈ z ответ: x = π/4 + πn, n ∈ z; arctg(-3) + πk, k ∈ z.
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