解答
tan(x+π)−cos(x+2π)=0
解答
x=2πn,x=π+2πn
+1
度数
x=0∘+360∘n,x=180∘+360∘n求解步骤
tan(x+π)−cos(x+2π)=0
使用三角恒等式改写
tan(x+π)−cos(x+2π)=0
使用三角恒等式改写
cos(x+2π)
使用角和恒等式: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(x)cos(2π)−sin(x)sin(2π)
化简 cos(x)cos(2π)−sin(x)sin(2π):−sin(x)
cos(x)cos(2π)−sin(x)sin(2π)
cos(x)cos(2π)=0
cos(x)cos(2π)
化简 cos(2π):0
cos(2π)
使用以下普通恒等式:cos(2π)=0
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=0=0⋅cos(x)
使用法则 0⋅a=0=0
sin(x)sin(2π)=sin(x)
sin(x)sin(2π)
化简 sin(2π):1
sin(2π)
使用以下普通恒等式:sin(2π)=1
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=1=1⋅sin(x)
乘以:sin(x)⋅1=sin(x)=sin(x)
=0−sin(x)
0−sin(x)=−sin(x)=−sin(x)
=−sin(x)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=cos(x+π)sin(x+π)
使用角和恒等式: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=cos(x+π)sin(x)cos(π)+cos(x)sin(π)
使用角和恒等式: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(x)cos(π)−sin(x)sin(π)sin(x)cos(π)+cos(x)sin(π)
化简 cos(x)cos(π)−sin(x)sin(π)sin(x)cos(π)+cos(x)sin(π):cos(x)sin(x)
cos(x)cos(π)−sin(x)sin(π)sin(x)cos(π)+cos(x)sin(π)
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)
=cos(π)cos(x)−sin(π)sin(x)−sin(x)
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)−sin(x)
使用分式法则: −b−a=ba=cos(x)sin(x)
=cos(x)sin(x)
cos(x)sin(x)−(−sin(x))=0
化简 cos(x)sin(x)−(−sin(x)):cos(x)sin(x)+sin(x)
cos(x)sin(x)−(−sin(x))
使用法则 −(−a)=a=cos(x)sin(x)+sin(x)
cos(x)sin(x)+sin(x)=0
cos(x)sin(x)+sin(x)=0
化简 cos(x)sin(x)+sin(x):cos(x)sin(x)+sin(x)cos(x)
cos(x)sin(x)+sin(x)
将项转换为分式: sin(x)=cos(x)sin(x)cos(x)=cos(x)sin(x)+cos(x)sin(x)cos(x)
因为分母相等，所以合并分式: ca±cb=ca±b=cos(x)sin(x)+sin(x)cos(x)
cos(x)sin(x)+sin(x)cos(x)=0
g(x)f(x)=0⇒f(x)=0sin(x)+sin(x)cos(x)=0
分解 sin(x)+sin(x)cos(x):sin(x)(cos(x)+1)
sin(x)+sin(x)cos(x)
因式分解出通项 sin(x)=sin(x)(1+cos(x))
sin(x)(cos(x)+1)=0
分别求解每个部分sin(x)=0orcos(x)+1=0
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
sin(x)=0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=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
解 x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
cos(x)+1=0:x=π+2πn
cos(x)+1=0
将 1到右边
cos(x)+1=0
两边减去 1cos(x)+1−1=0−1
化简cos(x)=−1
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
合并所有解x=2πn,x=π+2πn