это из ЭГЕ , задание N13 sqrt-cos2x+sqrt2*cosx=0 ответ надо расписать. Created by Im17ke. algebra-ru

(√2cos²x -cosx)*sqrt(-6sinx) =0 а)----------------------------------------1) sinx=0 ⇒ x₁ = π*k₁ ; k₁ ∈Z .2) {sinx ≤0; ( система ) {√2cos²x - cosx =0 ; √2cosx(cosx -1/√2) =0 ⇒ cosx = 0 и sinx ≤ 0 или cosx=1/√2 и sinx ≤ 0 x₂ = - π/2 +2π*k₂ ; x₃ = - π/4+2π*k₃ .---------------------------------------------------------------б) x ∈ [ 5π/2 ; 4π) .Из общих решений выбираем те которые принадлежат промежутку [ 5π/2 ; 4π) . k ₁ = 2 ;3 , k₂ =2 k₃ =2.3π ; 4π ; 7π/2 ; 15π/4.Новые вопросы в Алгебрапомогите пожалуйста решить методом сложения Помогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х÷х+9=14ПЛС М+ПОМОГИТЕ, ДАМС КОРОНУ И МНОГО 30 БАЛЛОВ(Нужно объяснение)557,559 номера. Если возможно,с решением Разложить на множетелиy^8-y^4+4y^2-4Дайте відповідь з рішенням будь ласка(557,559)1) 3 1/9 : 2 1/3 - 2 5/6 =28/9 * 3/7 - 17/6=4/3-17/6= 8/6-17/6=-3/2=-1.5как он получил -3/2Іван зірвав на клумбі 8 нарцисів та 4 тюльпани. Скільки всього існуєспособів вибору із цих квітів 3 нарцисів та 2 тюльпанів для букета?ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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это из ЭГЕ , задание N13 sqrt-cos2x+sqrt2*cosx=0 ответ надо расписать...

Im17ke @Im17ke

August 2022 1 1 Report

это из ЭГЕ , задание N13 sqrt-cos2x+sqrt2*cosx=0
ответ надо расписать

Answers & Comments

Sedenkov2017 (√2cos²x -cosx)*sqrt(-6sinx) =0
а)----------------------------------------
1) sinx=0 ⇒ x₁ = π*k₁ ; k₁ ∈Z .
2) {sinx ≤0; ( система )
   {√2cos²x - cosx =0 ;
√2cosx(cosx -1/√2) =0 ⇒ cosx = 0 и sinx ≤ 0    или cosx=1/√2 и sinx ≤ 0
x₂ = - π/2 +2π*k₂  ;
x₃ =  - π/4+2π*k₃ .
---------------------------------------------------------------
б) x ∈ [ 5π/2 ; 4π) .
Из общих решений выбираем те которые принадлежат промежутку
[ 5π/2 ; 4π) .
k ₁ = 2 ;3 , k₂ =2 k₃ =2.
3π ; 4π ; 7π/2 ; 15π/4.

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