解答
cos(x)=(2−tan(x))(1+sin(x))
解答
x=3π+2πn,x=35π+2πn
+1
度数
x=60∘+360∘n,x=300∘+360∘n求解步骤
cos(x)=(2−tan(x))(1+sin(x))
两边减去 (2−tan(x))(1+sin(x))cos(x)−(2−tan(x))(1+sin(x))=0
用 sin, cos 表示
cos(x)−(1+sin(x))(2−tan(x))
使用基本三角恒等式: tan(x)=cos(x)sin(x)=cos(x)−(1+sin(x))(2−cos(x)sin(x))
化简 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))
化简 2−cos(x)sin(x):cos(x)2cos(x)−sin(x)
2−cos(x)sin(x)
将项转换为分式: 2=cos(x)2cos(x)=cos(x)2cos(x)−cos(x)sin(x)
因为分母相等，所以合并分式: ca±cb=ca±b=cos(x)2cos(x)−sin(x)
=cos(x)2cos(x)−sin(x)(sin(x)+1)
分式相乘: a⋅cb=ca⋅b=cos(x)(2cos(x)−sin(x))(1+sin(x))
=cos(x)−cos(x)(2cos(x)−sin(x))(sin(x)+1)
将项转换为分式: cos(x)=cos(x)cos(x)cos(x)=cos(x)cos(x)cos(x)−cos(x)(2cos(x)−sin(x))(1+sin(x))
因为分母相等，所以合并分式: 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)
使用指数法则: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=cos1+1(x)
数字相加: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
使用三角恒等式改写
cos2(x)−(−sin(x)+2cos(x))(1+sin(x))
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=1−sin2(x)−(−sin(x)+2cos(x))(1+sin(x))
化简 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))
乘开 −(−sin(x)+2cos(x))(1+sin(x)):sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)
乘开 (−sin(x)+2cos(x))(1+sin(x)):−sin(x)−sin2(x)+2cos(x)+2cos(x)sin(x)
(−sin(x)+2cos(x))(1+sin(x))
使用 FOIL 方法: (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)
使用加减运算法则+(−a)=−a=−1⋅sin(x)−sin(x)sin(x)+2⋅1⋅cos(x)+2cos(x)sin(x)
化简 −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)
乘以:1⋅sin(x)=sin(x)=sin(x)
sin(x)sin(x)=sin2(x)
sin(x)sin(x)
使用指数法则: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=sin1+1(x)
数字相加:1+1=2=sin2(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1⋅cos(x)
数字相乘: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))
打开括号=−(−sin(x))−(−sin2(x))−(2cos(x))−(2cos(x)sin(x))
使用加减运算法则−(−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)
化简 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)
对同类项分组=−sin2(x)+sin(x)+sin2(x)−2cos(x)−2cos(x)sin(x)+1
同类项相加:−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
分解 1+sin(x)−2cos(x)−2cos(x)sin(x):(1−2cos(x))(sin(x)+1)
1+sin(x)−2cos(x)−2cos(x)sin(x)
因式分解出通项 sin(x)=1+sin(x)(1−2cos(x))−2cos(x)
改写为=(1−2cos(x))sin(x)+1⋅(1−2cos(x))
因式分解出通项 (1−2cos(x))=(1−2cos(x))(sin(x)+1)
(1−2cos(x))(sin(x)+1)=0
分别求解每个部分1−2cos(x)=0orsin(x)+1=0
1−2cos(x)=0:x=3π+2πn,x=35π+2πn
1−2cos(x)=0
将 1到右边
1−2cos(x)=0
两边减去 11−2cos(x)−1=0−1
化简−2cos(x)=−1
−2cos(x)=−1
两边除以 −2
−2cos(x)=−1
两边除以 −2−2−2cos(x)=−2−1
化简cos(x)=21
cos(x)=21
cos(x)=21的通解
cos(x) 周期表（周期为 2πn）：
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
将 1到右边
sin(x)+1=0
两边减去 1sin(x)+1−1=0−1
化简sin(x)=−1
sin(x)=−1
sin(x)=−1的通解
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
x=23π+2πn
x=23π+2πn
合并所有解x=3π+2πn,x=35π+2πn,x=23π+2πn
因为方程对以下值无定义:23π+2πnx=3π+2πn,x=35π+2πn