1)cos2x+2sin2x+2=0
(1 - tg²x)/(1 + tg²x) + 2*2tgx/(1 + tg²x) +2 = 0 | * (1 + tg²x )≠ 0
1 - tg²x + 4tgx +2(1 +tg²x) = 0
1 - tg²x + 4tgx +2 +2tg²x = 0
tg²x +4tgx +3 = 0
По т. Виета корни -1 и -3
а) tgx = -1 б) tgx = -3
x = -π/4 + πk, k ∈Z x = -arctg 3 + πn , n ∈Z
2)2cos2x+2sin²x=5+4sin2x
2(Cos²x - Sin²x) +2Sin²x = 5 +8SinxCosx
2Cos²x - 2Sin²x +2Sin²x = 5*1 +8SinxCosx
2Cos²x = 5*(Sin²x + Cos²x) + 10SinxCosx
2Cos²x = 5Sin²x + 5Cos²x +8SinxCosx
5Sin²x +3Cos²x +8sinxCosx = 0 |: Cos²x
5tg²x + 8tgx +3 = 0
tgx = t
5t² +8t +3 = 0
t = (-4 +-√(16 -15))/5 = (-5 +-1)/5
t₁ = -6/5= -1.2 t₂ = -4/5= -0,8
tgx =- 1,2 tgx = -0,8
x = -arctg1,2+ πk , k ∈ Z x = -arctg0,8 + πn , n ∈Z
3)2sin2x=3-2sin²x
6SinxCosx = 3Sin²x + 3Cos²x -2Sin²x
6SinxCosx = Sin²x + 3Cos²x | : Cos²x
6tgx = tg²x +3
tgx = t
t² -6t +3 = 0
t =3+-√(9 -3)
t₁ = 3 +√6 t₂ = 3 - √6
tgx = 3 +√6 tgx = 3 - √6
x = arctg(3 + √6) + πk , k ∈Z x = arctg(3 - √6) + πn , n ∈ Z