Question

Cos(/2+2x=√2sinx [-5; -4]

Answer 1

a)  cos( п/2 + 2х ) = √2sinx

- sin2x - √2sinx = 0

Применим формулу синуса двойного аргумента:  sin2x = 2sinx•cosx

2sinx•cosx + √2sinx = 0

sinx•( 2cosx + √2 ) = 0

1)  sinx = 0  ⇒  x = пn, n ∈ Z

2)  2cosx + √2 = 0  ⇒  cosx = - √2/2  ⇒  x = ± п/4 + 2пk, k ∈ Z

б)  С помощью тригонометрической окружности отберём корни, принадлежащие промежутку [  - 5п ; - 4п ] :

х₁ = - 5п

х₂ = - 4п - п/4 = - 17п/4

х₃ = - 4п

ОТВЕТ: а) пn, n ∈ Z ; ± п/4 + 2пk, k ∈ Z ; б) - 5п ; - 17п/4 ; - 4п

Answer 2

It seems there might be a typographical error or missing information in the equation you provided. The equation is not well-formed, and it's unclear what you are trying to solve for. Please double-check the equation and ensure it is written correctly. If you can provide the correct equation or specify what you want to solve for, I'd be happy to help you with the problem.