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stiklin
@stiklin
August 2022
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Помогите решить))
247(3)
248(1,3)
Большое спасибо!
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oganesbagoyan
Verified answer
Task/26532569
-------------------
tg(2x+π/3) =1 , (0; π/2) ;
2x +π
/
3 = π
/
4 + πn , n ∈
ℤ .
x = - π
/
24 +(π
/
2)*n , n ∈ ℤ .
n=1 ⇒ x = π
/
2 - π
/
24 =
11π/24
= 11*180°/24 =
82,5°
.
ответ :
82,5°
.
--------------------
2sinx +3cosx = 0 ;
2 sinx = -3cosx ;
* * * ясно cosx ≠ 0 иначе ⇒ sinx =0 ,но sin²x+ cos²x=1 * * *
tgx = -3/2 ;
x = - arctg(1,5) +πn ,n ∈ ℤ .
ответ :
- arctg(1,5) +πn ,n ∈ ℤ .
--------------------
sin²3x =cos²3x ;
(1 - cos6x) /2 =(1 + cos6x) /2 ;
1 - cos6x = 1 + cos6x ;
2cos6x = 0 ;
cos6x =0 ;
6x = π
/
2 + πn ,n ∈ ℤ .
x = π/12 +(π/6)*n ,n ∈ ℤ .
ответ :
π/12 +(π/6)*n ,n ∈ ℤ .
-----------------
Удачи !
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Answers & Comments
Verified answer
Task/26532569-------------------
tg(2x+π/3) =1 , (0; π/2) ;
2x +π/3 = π/4 + πn , n ∈ ℤ .
x = - π/24 +(π/2)*n , n ∈ ℤ .
n=1 ⇒ x = π/2 - π/24 = 11π/24 = 11*180°/24 = 82,5°.
ответ : 82,5° .
--------------------
2sinx +3cosx = 0 ;
2 sinx = -3cosx ;
* * * ясно cosx ≠ 0 иначе ⇒ sinx =0 ,но sin²x+ cos²x=1 * * *
tgx = -3/2 ;
x = - arctg(1,5) +πn ,n ∈ ℤ .
ответ : - arctg(1,5) +πn ,n ∈ ℤ .
--------------------
sin²3x =cos²3x ;
(1 - cos6x) /2 =(1 + cos6x) /2 ;
1 - cos6x = 1 + cos6x ;
2cos6x = 0 ;
cos6x =0 ;
6x = π/2 + πn ,n ∈ ℤ .
x = π/12 +(π/6)*n ,n ∈ ℤ .
ответ : π/12 +(π/6)*n ,n ∈ ℤ .
-----------------
Удачи !