解
cos(x)=(2−tan(x))(1+sin(x))
解
x=3π+2πn,x=35π+2πn
+1
度
x=60∘+360∘n,x=300∘+360∘n解答ステップ
cos(x)=(2−tan(x))(1+sin(x))
両辺から(2−tan(x))(1+sin(x))を引くcos(x)−(2−tan(x))(1+sin(x))=0
サイン, コサインで表わす
cos(x)−(1+sin(x))(2−tan(x))
基本的な三角関数の公式を使用する: tan(x)=cos(x)sin(x)=cos(x)−(1+sin(x))(2−cos(x)sin(x))
簡素化 cos(x)−(1+sin(x))(2−cos(x)sin(x)):cos(x)cos2(x)−(2cos(x)−sin(x))(1+sin(x))
cos(x)−(1+sin(x))(2−cos(x)sin(x))
(1+sin(x))(2−cos(x)sin(x))=cos(x)(2cos(x)−sin(x))(1+sin(x))
(1+sin(x))(2−cos(x)sin(x))
結合 2−cos(x)sin(x):cos(x)2cos(x)−sin(x)
2−cos(x)sin(x)
元を分数に変換する: 2=cos(x)2cos(x)=cos(x)2cos(x)−cos(x)sin(x)
分母が等しいので, 分数を組み合わせる: ca±cb=ca±b=cos(x)2cos(x)−sin(x)
=cos(x)2cos(x)−sin(x)(sin(x)+1)
分数を乗じる: a⋅cb=ca⋅b=cos(x)(2cos(x)−sin(x))(1+sin(x))
=cos(x)−cos(x)(2cos(x)−sin(x))(sin(x)+1)
元を分数に変換する: cos(x)=cos(x)cos(x)cos(x)=cos(x)cos(x)cos(x)−cos(x)(2cos(x)−sin(x))(1+sin(x))
分母が等しいので, 分数を組み合わせる: ca±cb=ca±b=cos(x)cos(x)cos(x)−(2cos(x)−sin(x))(1+sin(x))
cos(x)cos(x)−(2cos(x)−sin(x))(1+sin(x))=cos2(x)−(2cos(x)−sin(x))(1+sin(x))
cos(x)cos(x)−(2cos(x)−sin(x))(1+sin(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)
=cos2(x)−(2cos(x)−sin(x))(sin(x)+1)
=cos(x)cos2(x)−(2cos(x)−sin(x))(sin(x)+1)
=cos(x)cos2(x)−(2cos(x)−sin(x))(1+sin(x))
cos(x)cos2(x)−(−sin(x)+2cos(x))(1+sin(x))=0
g(x)f(x)=0⇒f(x)=0cos2(x)−(−sin(x)+2cos(x))(1+sin(x))=0
三角関数の公式を使用して書き換える
cos2(x)−(−sin(x)+2cos(x))(1+sin(x))
ピタゴラスの公式を使用する: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=1−sin2(x)−(−sin(x)+2cos(x))(1+sin(x))
簡素化 1−sin2(x)−(−sin(x)+2cos(x))(1+sin(x)):sin(x)−2cos(x)−2cos(x)sin(x)+1
1−sin2(x)−(−sin(x)+2cos(x))(1+sin(x))
拡張 −(−sin(x)+2cos(x))(1+sin(x)):sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)
拡張 (−sin(x)+2cos(x))(1+sin(x)):−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x)
(−sin(x)+2cos(x))(1+sin(x))
FOIL メソッドを適用する: (a+b)(c+d)=ac+ad+bc+bda=−sin(x),b=2cos(x),c=1,d=sin(x)=(−sin(x))⋅1+(−sin(x))sin(x)+2cos(x)⋅1+2cos(x)sin(x)
マイナス・プラスの規則を適用する+(−a)=−a=−1⋅sin(x)−sin(x)sin(x)+2⋅1⋅cos(x)+2cos(x)sin(x)
簡素化 −1⋅sin(x)−sin(x)sin(x)+2⋅1⋅cos(x)+2cos(x)sin(x):−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x)
−1⋅sin(x)−sin(x)sin(x)+2⋅1⋅cos(x)+2cos(x)sin(x)
1⋅sin(x)=sin(x)
1⋅sin(x)
乗算:1⋅sin(x)=sin(x)=sin(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)
2⋅1⋅cos(x)=2cos(x)
2⋅1⋅cos(x)
数を乗じる:2⋅1=2=2cos(x)
=−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x)
=−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x)
=−(−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x))
括弧を分配する=−(−sin(x))−(−sin2(x))−(2cos(x))−(2cos(x)sin(x))
マイナス・プラスの規則を適用する−(−a)=a,−(a)=−a=sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)
=1−sin2(x)+sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)
簡素化 1−sin2(x)+sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x):sin(x)−2cos(x)−2cos(x)sin(x)+1
1−sin2(x)+sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)
条件のようなグループ=−sin2(x)+sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)+1
類似した元を足す:−sin2(x)+sin2(x)=0=sin(x)−2cos(x)−2cos(x)sin(x)+1
=sin(x)−2cos(x)−2cos(x)sin(x)+1
=sin(x)−2cos(x)−2cos(x)sin(x)+1
1+sin(x)−2cos(x)−2cos(x)sin(x)=0
因数 1+sin(x)−2cos(x)−2cos(x)sin(x):(1−2cos(x))(sin(x)+1)
1+sin(x)−2cos(x)−2cos(x)sin(x)
共通項をくくり出す sin(x)=1+sin(x)(1−2cos(x))−2cos(x)
書き換え=(1−2cos(x))sin(x)+1⋅(1−2cos(x))
共通項をくくり出す (1−2cos(x))=(1−2cos(x))(sin(x)+1)
(1−2cos(x))(sin(x)+1)=0
各部分を別個に解く1−2cos(x)=0orsin(x)+1=0
1−2cos(x)=0:x=3π+2πn,x=35π+2πn
1−2cos(x)=0
1を右側に移動します
1−2cos(x)=0
両辺から1を引く1−2cos(x)−1=0−1
簡素化−2cos(x)=−1
−2cos(x)=−1
以下で両辺を割る−2
−2cos(x)=−1
以下で両辺を割る−2−2−2cos(x)=−2−1
簡素化cos(x)=21
cos(x)=21
以下の一般解 cos(x)=21
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=3π+2πn,x=35π+2πn
x=3π+2πn,x=35π+2πn
sin(x)+1=0:x=23π+2πn
sin(x)+1=0
1を右側に移動します
sin(x)+1=0
両辺から1を引くsin(x)+1−1=0−1
簡素化sin(x)=−1
sin(x)=−1
以下の一般解 sin(x)=−1
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=23π+2πn
x=23π+2πn
すべての解を組み合わせるx=3π+2πn,x=35π+2πn,x=23π+2πn
equationは以下で未定義のため:23π+2πnx=3π+2πn,x=35π+2πn