Question

Помогите решить тригонометрические уравнения:
1) sin(x+4π/3)=2sin(4π/3-x)
2) корень из ((1-cosx)^2+sin^2x)=2sinx/2
3) 5sin2x+5cosx-8sinx-4=0

Answer 1

 √3/2 Cosx +1/2Sinx 1)Sin(π  + π/3+x) = 2Sin(π+π/3 - x)
   -Sin(π/3 +х) = -2Sin(π/3 -x)
   Sin(π/3 +х) = 2Sin(π/3 -x)
   Sin π/3Cosx + Cosπ/3Sinx = 2(Sin π/3Cosx - Cosπ/3Sinx )
   √3/2 Cosx + 1/2Sinx  = 2( √3/2 Cosx -1/2Sinx )
   √3/2 Cosx +1/2Sinx =  √3 Cosx - Sinx 
  √3/2 Cosx - √3Cosx  +1/2Sinx + Sinx = 0
-√3Cosx + 3/2Sinx = 0
3/2Sinx = √3Cosx | : 3/2Cosx
tgx = 2√3/3 
x = arctg2√3/3 + πk , k ∈Z
2)√(1 - 2Cosx + Cos²x + Sin²x) = 2Sinx/2
    √(1 - 2Cosx +1) = 2Sinx/2
    √(2-2Cosx) = 2Sinx/2
    √2(1 - Cosx) = 2Sinx/2
   √4(1 - Cosx)/2 = 2Sinx/2
2√(1-Сosx)/2= 2Sinx/2
 +- Sinx/2 = Sinx/2
2Sinx/2 = 0
Sinx/2 = 0
x/2 = πn, n ∈ Z
x = 2πn, n ∈ Z
3) 5*2SinxCosx + 5Cosx -8Sinx -4= 0
10SinxCosx +5Cosx -8Sinx -4 = 0
5Cosx(2Sinx +1) -4(2Sinx +1) = 0
(2Sinx +1)(5Cosx -4) = 0
2Sinx +1 = 0            или             5Cosx -4 = 0
a) Sinx = -1/2                              б) Cosx = 4/5
x = (-1)ⁿ⁺¹ π/6 + nπ, n ∈Z                x = +-arcCos4/5 + 2πk, k ∈ Z