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
5^(x-2) = 5^0 2^(x² -3x +8) = 2^6
x-2 = 0 x² -3x +8 = 6
x = 2 x² -3x +2 = 0
2) 3·4^x =48 x = 1 и х = 2
4^x = 16 6)7^(2x-8)·7^(x+7) = 0
4^x = 4² нет решений
x=2 7)(0,2)^x ≤ 25·5√5
3)3^x=27·3√9 5^-x ≤ 5²·5·5^1/2
3^x = 3³·3·3 5^-x ≤5^3,5
3^x = 3^5 -x ≤ 3,5
x = 5 x ≥ -3,5
4)3^x + 3^(x +1) = 4 8)(1/2)^-x + 2^(3 +x) ≤9
3^x(1 +3) = 4 2^x +2^(3 +x) ≤ 9
3^x·4 = 4 2^x(1 +2^3) ≤ 9 | :9
3^x = 1 2^x ≤ 1
x = 0 2^x ≤2^0
x≤ 0