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AngelSkyOO
@AngelSkyOO
July 2022
1
2
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1) 6sin^2(x)+15sin(3¶/2+x)-12=0, [-5¶; - 7¶/2]
2) 2cos(x-11¶/2)*cosx=sinx, [-5¶; - 4¶]
3) cosx=(cos(x/2)-sin(x/2))^2 - 1
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sedinalana
Verified answer
1
6sin²x-15cosx-12=0
6-6cos²x-15cosx-12=0
cosx=a
6a²+15a+6=0
2a²+5a+2=0
D=25-16=9
a1=(-5-3)/4=-2⇒cosx=-2<-1 нет решения
a2=(-5+3)/4=-1/2⇒cosx=-1/2
x1=-2π/3+2πn,n∈z U x2=2π/3+2πk,k∈z
-5π≤-2π/3+2πn≤-7π/2
-30≤-4+12n≤-21
-26≤12n≤-17
-13/6≤n≤-17/12
n=-2⇒x=-2π/3-4π=-14π/3
-5π≤2π/3+2πk≤-7π/2
-30≤4+12k≤-21
-34≤12k≤-25
-17/6≤k≤-25/12
нет решения
Ответ х=-14π/3
2
2sinx*cosx=sinx
2sinxcosx-sinx=0
sinx*(2cosx-1)=0
sinx=0⇒x1=πn,n∈z
-5π≤πn≤-4π
-5≤n≤-4
n=-5⇒x=-5π
n=-4⇒x=-4π
cosx=1/2⇒x2=-π/3 +2πk,k∈z U x3=π/3+2πt,t∈z
-5π≤-π/3+2πk≤-4π
-15≤-1+6k≤-12
-14≤6k≤-11
-7/3≤k≤-11/6
k=-2⇒x=-π/3-4π=-13π/3
-5π≤π/3+2πt≤-4π
-15≤1+6t≤-12
-16≤6t≤-13
-8/3≤t≤-13/6
нет решения
Ответ x={-5π;-4π;-13π/3}
3
cosx=cos²x/2-2cosx/2*sinx/2+sin²x/2-1
cosx=1-sinx-1
cosx=-sinx
cosx+sinx=0/cosx
1+tgx=0
tgx=-1⇒x=-π/4+πn
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Answers & Comments
Verified answer
16sin²x-15cosx-12=0
6-6cos²x-15cosx-12=0
cosx=a
6a²+15a+6=0
2a²+5a+2=0
D=25-16=9
a1=(-5-3)/4=-2⇒cosx=-2<-1 нет решения
a2=(-5+3)/4=-1/2⇒cosx=-1/2
x1=-2π/3+2πn,n∈z U x2=2π/3+2πk,k∈z
-5π≤-2π/3+2πn≤-7π/2
-30≤-4+12n≤-21
-26≤12n≤-17
-13/6≤n≤-17/12
n=-2⇒x=-2π/3-4π=-14π/3
-5π≤2π/3+2πk≤-7π/2
-30≤4+12k≤-21
-34≤12k≤-25
-17/6≤k≤-25/12
нет решения
Ответ х=-14π/3
2
2sinx*cosx=sinx
2sinxcosx-sinx=0
sinx*(2cosx-1)=0
sinx=0⇒x1=πn,n∈z
-5π≤πn≤-4π
-5≤n≤-4
n=-5⇒x=-5π
n=-4⇒x=-4π
cosx=1/2⇒x2=-π/3 +2πk,k∈z U x3=π/3+2πt,t∈z
-5π≤-π/3+2πk≤-4π
-15≤-1+6k≤-12
-14≤6k≤-11
-7/3≤k≤-11/6
k=-2⇒x=-π/3-4π=-13π/3
-5π≤π/3+2πt≤-4π
-15≤1+6t≤-12
-16≤6t≤-13
-8/3≤t≤-13/6
нет решения
Ответ x={-5π;-4π;-13π/3}
3
cosx=cos²x/2-2cosx/2*sinx/2+sin²x/2-1
cosx=1-sinx-1
cosx=-sinx
cosx+sinx=0/cosx
1+tgx=0
tgx=-1⇒x=-π/4+πn