解
1+cos(x)sin(x)+cot(x)=2
解
x=6π+2πn,x=65π+2πn
+1
度
x=30∘+360∘n,x=150∘+360∘n解答ステップ
1+cos(x)sin(x)+cot(x)=2
両辺から2を引く1+cos(x)sin(x)+cot(x)−2=0
簡素化 1+cos(x)sin(x)+cot(x)−2:1+cos(x)sin(x)+cot(x)(1+cos(x))−2(1+cos(x))
1+cos(x)sin(x)+cot(x)−2
元を分数に変換する: cot(x)=1+cos(x)cot(x)(1+cos(x)),2=1+cos(x)2(1+cos(x))=1+cos(x)sin(x)+1+cos(x)cot(x)(1+cos(x))−1+cos(x)2(1+cos(x))
分母が等しいので, 分数を組み合わせる: ca±cb=ca±b=1+cos(x)sin(x)+cot(x)(1+cos(x))−2(1+cos(x))
1+cos(x)sin(x)+cot(x)(1+cos(x))−2(1+cos(x))=0
g(x)f(x)=0⇒f(x)=0sin(x)+cot(x)(1+cos(x))−2(1+cos(x))=0
サイン, コサインで表わす
sin(x)−(1+cos(x))⋅2+(1+cos(x))cot(x)
基本的な三角関数の公式を使用する: cot(x)=sin(x)cos(x)=sin(x)−(1+cos(x))⋅2+(1+cos(x))sin(x)cos(x)
簡素化 sin(x)−(1+cos(x))⋅2+(1+cos(x))sin(x)cos(x):sin(x)sin2(x)−2sin(x)(1+cos(x))+cos(x)(1+cos(x))
sin(x)−(1+cos(x))⋅2+(1+cos(x))sin(x)cos(x)
乗じる (1+cos(x))sin(x)cos(x):sin(x)cos(x)(cos(x)+1)
(1+cos(x))sin(x)cos(x)
分数を乗じる: a⋅cb=ca⋅b=sin(x)cos(x)(1+cos(x))
=sin(x)−2(cos(x)+1)+sin(x)cos(x)(cos(x)+1)
元を分数に変換する: sin(x)=sin(x)sin(x)sin(x),2(cos(x)+1)=sin(x)(1+cos(x))2sin(x)=sin(x)sin(x)sin(x)−sin(x)(1+cos(x))⋅2sin(x)+sin(x)cos(x)(1+cos(x))
分母が等しいので, 分数を組み合わせる: ca±cb=ca±b=sin(x)sin(x)sin(x)−(1+cos(x))⋅2sin(x)+cos(x)(1+cos(x))
sin(x)sin(x)−(1+cos(x))⋅2sin(x)+cos(x)(1+cos(x))=sin2(x)−2sin(x)(1+cos(x))+cos(x)(1+cos(x))
sin(x)sin(x)−(1+cos(x))⋅2sin(x)+cos(x)(1+cos(x))
sin(x)sin(x)=sin2(x)
sin(x)sin(x)
指数の規則を適用する: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=sin1+1(x)
数を足す:1+1=2=sin2(x)
=sin2(x)−2sin(x)(cos(x)+1)+cos(x)(cos(x)+1)
=sin(x)sin2(x)−2sin(x)(cos(x)+1)+cos(x)(cos(x)+1)
=sin(x)sin2(x)−2sin(x)(1+cos(x))+cos(x)(1+cos(x))
sin(x)sin2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x)=0
g(x)f(x)=0⇒f(x)=0sin2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x)=0
三角関数の公式を使用して書き換える
sin2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x)
ピタゴラスの公式を使用する: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=1−cos2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x)
簡素化 1−cos2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x):cos(x)−2sin(x)−2sin(x)cos(x)+1
1−cos2(x)+(1+cos(x))cos(x)−(1+cos(x))⋅2sin(x)
=1−cos2(x)+cos(x)(1+cos(x))−2sin(x)(1+cos(x))
拡張 cos(x)(1+cos(x)):cos(x)+cos2(x)
cos(x)(1+cos(x))
分配法則を適用する: a(b+c)=ab+aca=cos(x),b=1,c=cos(x)=cos(x)⋅1+cos(x)cos(x)
=1⋅cos(x)+cos(x)cos(x)
簡素化 1⋅cos(x)+cos(x)cos(x):cos(x)+cos2(x)
1⋅cos(x)+cos(x)cos(x)
1⋅cos(x)=cos(x)
1⋅cos(x)
乗算:1⋅cos(x)=cos(x)=cos(x)
cos(x)cos(x)=cos2(x)
cos(x)cos(x)
指数の規則を適用する: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=cos1+1(x)
数を足す:1+1=2=cos2(x)
=cos(x)+cos2(x)
=cos(x)+cos2(x)
=1−cos2(x)+cos(x)+cos2(x)−(1+cos(x))⋅2sin(x)
拡張 −2sin(x)(1+cos(x)):−2sin(x)−2sin(x)cos(x)
−2sin(x)(1+cos(x))
分配法則を適用する: a(b+c)=ab+aca=−2sin(x),b=1,c=cos(x)=−2sin(x)⋅1+(−2sin(x))cos(x)
マイナス・プラスの規則を適用する+(−a)=−a=−2⋅1⋅sin(x)−2sin(x)cos(x)
数を乗じる:2⋅1=2=−2sin(x)−2sin(x)cos(x)
=1−cos2(x)+cos(x)+cos2(x)−2sin(x)−2sin(x)cos(x)
簡素化 1−cos2(x)+cos(x)+cos2(x)−2sin(x)−2sin(x)cos(x):cos(x)−2sin(x)−2sin(x)cos(x)+1
1−cos2(x)+cos(x)+cos2(x)−2sin(x)−2sin(x)cos(x)
条件のようなグループ=−cos2(x)+cos(x)+cos2(x)−2sin(x)−2sin(x)cos(x)+1
類似した元を足す:−cos2(x)+cos2(x)=0=cos(x)−2sin(x)−2sin(x)cos(x)+1
=cos(x)−2sin(x)−2sin(x)cos(x)+1
=cos(x)−2sin(x)−2sin(x)cos(x)+1
1+cos(x)−2sin(x)−2cos(x)sin(x)=0
因数 1+cos(x)−2sin(x)−2cos(x)sin(x):(1−2sin(x))(cos(x)+1)
1+cos(x)−2sin(x)−2cos(x)sin(x)
共通項をくくり出す cos(x)=1+cos(x)(1−2sin(x))−2sin(x)
書き換え=(1−2sin(x))cos(x)+1⋅(1−2sin(x))
共通項をくくり出す (1−2sin(x))=(1−2sin(x))(cos(x)+1)
(1−2sin(x))(cos(x)+1)=0
各部分を別個に解く1−2sin(x)=0orcos(x)+1=0
1−2sin(x)=0:x=6π+2πn,x=65π+2πn
1−2sin(x)=0
1を右側に移動します
1−2sin(x)=0
両辺から1を引く1−2sin(x)−1=0−1
簡素化−2sin(x)=−1
−2sin(x)=−1
以下で両辺を割る−2
−2sin(x)=−1
以下で両辺を割る−2−2−2sin(x)=−2−1
簡素化sin(x)=21
sin(x)=21
以下の一般解 sin(x)=21
sin(x)2πn 循環を含む周期性テーブル:
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
x=6π+2πn,x=65π+2πn
x=6π+2πn,x=65π+2πn
cos(x)+1=0:x=π+2πn
cos(x)+1=0
1を右側に移動します
cos(x)+1=0
両辺から1を引くcos(x)+1−1=0−1
簡素化cos(x)=−1
cos(x)=−1
以下の一般解 cos(x)=−1
cos(x)2πn 循環を含む周期性テーブル:
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=π+2πn
x=π+2πn
すべての解を組み合わせるx=6π+2πn,x=65π+2πn,x=π+2πn
equationは以下で未定義のため:π+2πnx=6π+2πn,x=65π+2πn