Question

Решить уравнения....помогите...) 1. sin2x=√2 cosx 2. sin2x+√2sinx=0

Answer 1

1) SIn2x = 2SinxCosx

   Sin2x = √2Cosx

   2SinxCosx - √2Cosx = 0

  Cosx(2Sinx - √2) = 0

Cos x = 0 или Sinx = √2/2

x₁ = π/2 + πk, k∈Z 

$x_2 = (-1)^k frac{pi}{4}+pi k$

 

2) 2SinxCosx + √2Sinx = 0

   Sinx(2Cosx + √2) = 0

Sinx = 0 или Cosx = -√2/2

x₁ = πk

x₂ = ±3π/4 + 2πk