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рау01
@рау01
July 2022
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2cos(p-2x)>1 помогите пожалуйста
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paradiseva
Verified answer
2cos(p-2x)>1
cos(pi - 2x) > 1/2
2pk
- arccos
1/2 < pi - 2x < 2pk
+ arccos
1/2, k∈Z
2pk - pi/3 < pi - 2x < 2pk + pi/3, k∈Z
2pk - pi/3 -pi < - 2x < 2pk + pi/3 - pi, k∈Z
2pk - 4pi/3 < - 2x < 2pk - 2pi/3, k∈Z
-pk + 2pi/3 > x > -pk + pi/3, k∈Z
pi/3 - pik < x < 2pi/3 - pik, k∈Z
3 votes
Thanks 4
oganesbagoyan
Verified answer
task/24804493
---.---.---.---.---.---
решить неравенство 2cos(
π
-2x) >1 ;
-2cos2x > 1 ;
cos2x < -1/2 ;
(
π -π/3) <2x < π+π/3 ;
2π/3 <2x < 4π/3 ;
2πn +2π/3 <2x < 4π/3 +2πn ;
πn +π/3 <x < 2π/3 +πn ; n ∈ Z * * * Z _множество целых чисел * * *
ответ : ∪ ( πn +π/3 ;
2
π/3 +
πn )
n ∈ Z .
* * * * * * * * * * * * * * * * * * * * * *
k = -n
∈ Z
0 votes
Thanks 2
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Answers & Comments
Verified answer
2cos(p-2x)>1cos(pi - 2x) > 1/2
2pk - arccos1/2 < pi - 2x < 2pk + arccos1/2, k∈Z
2pk - pi/3 < pi - 2x < 2pk + pi/3, k∈Z
2pk - pi/3 -pi < - 2x < 2pk + pi/3 - pi, k∈Z
2pk - 4pi/3 < - 2x < 2pk - 2pi/3, k∈Z
-pk + 2pi/3 > x > -pk + pi/3, k∈Z
pi/3 - pik < x < 2pi/3 - pik, k∈Z
Verified answer
task/24804493---.---.---.---.---.---
решить неравенство 2cos(π-2x) >1 ;
-2cos2x > 1 ;
cos2x < -1/2 ;
(π -π/3) <2x < π+π/3 ;
2π/3 <2x < 4π/3 ;
2πn +2π/3 <2x < 4π/3 +2πn ;
πn +π/3 <x < 2π/3 +πn ; n ∈ Z * * * Z _множество целых чисел * * *
ответ : ∪ ( πn +π/3 ; 2π/3 +πn )
n ∈ Z .
* * * * * * * * * * * * * * * * * * * * * *
k = -n ∈ Z