cos2x+√2cos(pi/2+x)+1=0 [2pi;7pi/2]

cos2x +√2 *cos(π/2 +x) +1 =0 ; x ∈[2π; 3,5π]. --------------------------------------------------- 1- 2sin²x +√2*(-sinx ) +1 = 0 ; 2sin²x +√2sinx -2 = 0  ;  * * * sin²x +(√2)/2sinx -1 = 0  * * * sinx = -√2    * * *-√2 < -1 не имеет решения * * * sinx =(√2)/2 .    * * * * * x =(-1)^n*π/4 +π*n , n∈Z   * * * * * x₁ = π/4 +2πn ⇒ x =π/4 +2π =9π/4 ∈[2π; 3,5π].. x₂ =π -π/4 +2πn = 3π/4 +2πn  ⇒x = 11π/4 ∈ [2π; 3,5π] . ответ : 9π/4 ;.11π/4 .