解答
csc(x−π)cos2(x−π)−cot(x−π)=0
解答
x=2π+2πn,x=23π+2πn
+1
度数
x=90∘+360∘n,x=270∘+360∘n求解步骤
csc(x−π)cos2(x−π)−cot(x−π)=0
使用三角恒等式改写
csc(x−π)cos2(x−π)−cot(x−π)=0
使用三角恒等式改写
cos(x−π)
使用角差恒等式: cos(s−t)=cos(s)cos(t)+sin(s)sin(t)=cos(x)cos(π)+sin(x)sin(π)
化简 cos(x)cos(π)+sin(x)sin(π):−cos(x)
cos(x)cos(π)+sin(x)sin(π)
cos(x)cos(π)=−cos(x)
cos(x)cos(π)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=(−1)cos(x)
整理后得=−cos(x)
=−cos(x)+sin(π)sin(x)
sin(x)sin(π)=0
sin(x)sin(π)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅sin(x)
使用法则 0⋅a=0=0
=−cos(x)+0
−cos(x)+0=−cos(x)=−cos(x)
=−cos(x)
使用基本三角恒等式: csc(x)=sin(x)1=sin(x−π)1
使用角差恒等式: sin(s−t)=sin(s)cos(t)−cos(s)sin(t)=sin(x)cos(π)−cos(x)sin(π)1
化简 sin(x)cos(π)−cos(x)sin(π)1:−sin(x)1
sin(x)cos(π)−cos(x)sin(π)1
sin(x)cos(π)−cos(x)sin(π)=−sin(x)
sin(x)cos(π)−cos(x)sin(π)
sin(x)cos(π)=−sin(x)
sin(x)cos(π)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=(−1)sin(x)
整理后得=−sin(x)
=−sin(x)−sin(π)cos(x)
cos(x)sin(π)=0
cos(x)sin(π)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅cos(x)
使用法则 0⋅a=0=0
=−sin(x)−0
−sin(x)−0=−sin(x)=−sin(x)
=−sin(x)1
使用分式法则: −ba=−ba=−sin(x)1
=−sin(x)1
使用基本三角恒等式: cot(x)=sin(x)cos(x)=sin(x−π)cos(x−π)
使用角差恒等式: sin(s−t)=sin(s)cos(t)−cos(s)sin(t)=sin(x)cos(π)−cos(x)sin(π)cos(x−π)
使用角差恒等式: cos(s−t)=cos(s)cos(t)+sin(s)sin(t)=sin(x)cos(π)−cos(x)sin(π)cos(x)cos(π)+sin(x)sin(π)
化简 sin(x)cos(π)−cos(x)sin(π)cos(x)cos(π)+sin(x)sin(π):sin(x)cos(x)
sin(x)cos(π)−cos(x)sin(π)cos(x)cos(π)+sin(x)sin(π)
cos(x)cos(π)+sin(x)sin(π)=−cos(x)
cos(x)cos(π)+sin(x)sin(π)
cos(x)cos(π)=−cos(x)
cos(x)cos(π)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=(−1)cos(x)
整理后得=−cos(x)
=−cos(x)+sin(π)sin(x)
sin(x)sin(π)=0
sin(x)sin(π)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅sin(x)
使用法则 0⋅a=0=0
=−cos(x)+0
−cos(x)+0=−cos(x)=−cos(x)
=cos(π)sin(x)−sin(π)cos(x)−cos(x)
sin(x)cos(π)−cos(x)sin(π)=−sin(x)
sin(x)cos(π)−cos(x)sin(π)
sin(x)cos(π)=−sin(x)
sin(x)cos(π)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=(−1)sin(x)
整理后得=−sin(x)
=−sin(x)−sin(π)cos(x)
cos(x)sin(π)=0
cos(x)sin(π)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅cos(x)
使用法则 0⋅a=0=0
=−sin(x)−0
−sin(x)−0=−sin(x)=−sin(x)
=−sin(x)−cos(x)
使用分式法则: −b−a=ba=sin(x)cos(x)
=sin(x)cos(x)
(−sin(x)1)(−cos(x))2−sin(x)cos(x)=0
化简 (−sin(x)1)(−cos(x))2−sin(x)cos(x):sin(x)−cos2(x)−cos(x)
(−sin(x)1)(−cos(x))2−sin(x)cos(x)
去除括号: (−a)=−a=−sin(x)1(−cos(x))2−sin(x)cos(x)
sin(x)1(−cos(x))2=sin(x)cos2(x)
sin(x)1(−cos(x))2
(−cos(x))2=cos2(x)
(−cos(x))2
使用指数法则: (−a)n=an,若 n 是偶数(−cos(x))2=cos2(x)=cos2(x)
=sin(x)1cos2(x)
分式相乘: a⋅cb=ca⋅b=sin(x)1⋅cos2(x)
乘以:1⋅cos2(x)=cos2(x)=sin(x)cos2(x)
=−sin(x)cos2(x)−sin(x)cos(x)
使用法则 ca±cb=ca±b=sin(x)−cos2(x)−cos(x)
sin(x)−cos2(x)−cos(x)=0
sin(x)−cos2(x)−cos(x)=0
g(x)f(x)=0⇒f(x)=0−cos2(x)−cos(x)=0
用替代法求解
−cos2(x)−cos(x)=0
令:cos(x)=u−u2−u=0
−u2−u=0:u=−1,u=0
−u2−u=0
使用求根公式求解
−u2−u=0
二次方程求根公式:
若 a=−1,b=−1,c=0u1,2=2(−1)−(−1)±(−1)2−4(−1)⋅0
u1,2=2(−1)−(−1)±(−1)2−4(−1)⋅0
(−1)2−4(−1)⋅0=1
(−1)2−4(−1)⋅0
使用法则 −(−a)=a=(−1)2+4⋅1⋅0
(−1)2=1
(−1)2
使用指数法则: (−a)n=an,若 n 是偶数(−1)2=12=12
使用法则 1a=1=1
4⋅1⋅0=0
4⋅1⋅0
使用法则 0⋅a=0=0
=1+0
数字相加:1+0=1=1
使用法则 1=1=1
u1,2=2(−1)−(−1)±1
将解分隔开u1=2(−1)−(−1)+1,u2=2(−1)−(−1)−1
u=2(−1)−(−1)+1:−1
2(−1)−(−1)+1
去除括号: (−a)=−a,−(−a)=a=−2⋅11+1
数字相加:1+1=2=−2⋅12
数字相乘:2⋅1=2=−22
使用分式法则: −ba=−ba=−22
使用法则 aa=1=−1
u=2(−1)−(−1)−1:0
2(−1)−(−1)−1
去除括号: (−a)=−a,−(−a)=a=−2⋅11−1
数字相减:1−1=0=−2⋅10
数字相乘:2⋅1=2=−20
使用分式法则: −ba=−ba=−20
使用法则 a0=0,a=0=−0
=0
二次方程组的解是:u=−1,u=0
u=cos(x)代回cos(x)=−1,cos(x)=0
cos(x)=−1,cos(x)=0
cos(x)=−1:x=π+2πn
cos(x)=−1
cos(x)=−1的通解
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=π+2πn
x=π+2πn
cos(x)=0:x=2π+2πn,x=23π+2πn
cos(x)=0
cos(x)=0的通解
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=2π+2πn,x=23π+2πn
x=2π+2πn,x=23π+2πn
合并所有解x=π+2πn,x=2π+2πn,x=23π+2πn
因为方程对以下值无定义:π+2πnx=2π+2πn,x=23π+2πn