task/29538500
356. 3)
sin2x +4(sinx+cosx)+4 =0⇔ 1+2sinxcosx+4(sinx+cosx) + 3 =0 ⇔
(sin²x+cos²x+2sinxcosx ) + 4(sinx+cosx) + 3 =0 ⇔
(sinx+cosx)²+ 4(sinx+cosx) + 3 =0 замена : t =sinx+cosx
t² +4t +3 =0 ⇔ [t = -1 ; t = - 3 . ⇒ [ sinx+cosx = -1 ; sinx+cosx = -3. ⇒
* * * sinx + cosx = -3 не имеет решения , т .к. -1 ≤ sinx ≤ 1 , -1 ≤ cosx ≤ 1 * * *
sinx + cosx = - 1 ⇔ √2sin (x +π/4) = -1 ⇔ sin (x +π/4) = -1/√2 ⇒
* * * x + π/4 = (-1)ⁿ⁺¹π/4 + πn , n∈ ℤ * * *
x + π/4 = π+ π/4 +2πn , n∈ ℤ ИЛИ x + π/4 = - π/4 +2πn , n∈ ℤ .
x = π+2πn , n∈ ℤ ИЛИ x = - π/2 +2πn , n∈ ℤ .
ответ : π + 2πn , - π/2 +2πn , n∈ ℤ .