znanija.com/task/34452584
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Решить уравнение√2cos(8x)cos(x+π/4)= 2cos(π /4)
Ответ: x = - π/4+2πk , k∈ ℤ .
Объяснение:
√2cos(8x)cos(x+π/4)= 2cos(π /4) ⇔√2cos(8x)cos(x+π/4)= 2*√2/2⇔
cos(8x)cos(x+π/4)= 1 .
a) { cos(8x)= -1 ; cos(x+π/4) = -1 .
{ 8x =π+2πn ; x+π/4 =π+2πk.⇔{ x =π/8+(π/4)*n ; x=3π/4+2πk. (n , k ∈ℤ)
⇒ x ∈ ∅ .
* * *π/8+(π/4)*n=3π/4+2πk ⇔1+2n =6+8k⇔ n - 4k =2,5 невозможно * * *
b) { cos(8x)= 1 ; cos(x+π/4) =1 .
{ 8x =2πn ; x+π/4 =2πk. (n , k ∈ℤ)
{ x =(π/4)*n ; x= - π/4+2πk. ⇒ x = - π/4+2πk k∈ ℤ .
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znanija.com/task/34452584
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Решить уравнение√2cos(8x)cos(x+π/4)= 2cos(π /4)
Ответ: x = - π/4+2πk , k∈ ℤ .
Объяснение:
√2cos(8x)cos(x+π/4)= 2cos(π /4) ⇔√2cos(8x)cos(x+π/4)= 2*√2/2⇔
cos(8x)cos(x+π/4)= 1 .
a) { cos(8x)= -1 ; cos(x+π/4) = -1 .
{ 8x =π+2πn ; x+π/4 =π+2πk.⇔{ x =π/8+(π/4)*n ; x=3π/4+2πk. (n , k ∈ℤ)
⇒ x ∈ ∅ .
* * *π/8+(π/4)*n=3π/4+2πk ⇔1+2n =6+8k⇔ n - 4k =2,5 невозможно * * *
b) { cos(8x)= 1 ; cos(x+π/4) =1 .
{ 8x =2πn ; x+π/4 =2πk. (n , k ∈ℤ)
{ x =(π/4)*n ; x= - π/4+2πk. ⇒ x = - π/4+2πk k∈ ℤ .