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Dianakisa
@Dianakisa
August 2022
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Решите уравнение sinx + cos2x=1 + sinx * cos2x
Найдите все корни этого уравнения, принадлежащие отрезку (0, П)
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sedinalana
Sinx+cos2x-1-sinxcos2x=0
(sinx-1)-cos2x(sinx-1)=0
(sinx-1)(1-cos2x)=0
sinx-1=0⇒sinx=1⇒x=π/2+2πk.k∈z
0≤π/2+2πk≤π
0≤1+4k≤2
-1≤4k≤1
-1/4≤k≤1/4
k=0πx=π/2
1-cos2x=0⇒cos2x=1⇒2x=2πk⇒x=πk,k∈z
0≤πk≤π
0≤k≤1
k=0⇒x=0
k=1⇒x=π
Ответ x={0;π/2;π}
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Answers & Comments
(sinx-1)-cos2x(sinx-1)=0
(sinx-1)(1-cos2x)=0
sinx-1=0⇒sinx=1⇒x=π/2+2πk.k∈z
0≤π/2+2πk≤π
0≤1+4k≤2
-1≤4k≤1
-1/4≤k≤1/4
k=0πx=π/2
1-cos2x=0⇒cos2x=1⇒2x=2πk⇒x=πk,k∈z
0≤πk≤π
0≤k≤1
k=0⇒x=0
k=1⇒x=π
Ответ x={0;π/2;π}