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alexdumler99
@alexdumler99
July 2022
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2cos²x+cos x-1=0 , x принадлежит [-7П/2 ; 2П)
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sedinalana
Verified answer
Cosx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosa=-1⇒a=π+2πn,n∈z
-7π/2≤π+2πn≤2π
-7≤2+4n≤4
-9≤4n≤2
-9/4≤n≤1/2
n=-2⇒a=π-4π=-3π
n=-1⇒a=π-2π=-π
n=0⇒a=π
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=π/3+2πk U x=-π/3+2πm
-7π/2≤π/3+2πk≤2π
-21≤2+12k≤12
-23≤12k≤10
-23/12≤k≤5/6
k=-1⇒a=π/3-2π=-5π/3
k=0⇒a=π/3
-7π/2≤-π/3+2πm≤2π
-21≤-2+12m≤12
-19≤12m≤14
-19/12≤m≤7/6
m=-1⇒a=-π/3-2π=-7π/3
m=0⇒a=-π/3
m=1⇒a=-π/3+2π=5π/3
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Answers & Comments
Verified answer
Cosx=a2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosa=-1⇒a=π+2πn,n∈z
-7π/2≤π+2πn≤2π
-7≤2+4n≤4
-9≤4n≤2
-9/4≤n≤1/2
n=-2⇒a=π-4π=-3π
n=-1⇒a=π-2π=-π
n=0⇒a=π
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=π/3+2πk U x=-π/3+2πm
-7π/2≤π/3+2πk≤2π
-21≤2+12k≤12
-23≤12k≤10
-23/12≤k≤5/6
k=-1⇒a=π/3-2π=-5π/3
k=0⇒a=π/3
-7π/2≤-π/3+2πm≤2π
-21≤-2+12m≤12
-19≤12m≤14
-19/12≤m≤7/6
m=-1⇒a=-π/3-2π=-7π/3
m=0⇒a=-π/3
m=1⇒a=-π/3+2π=5π/3