解答
tan(2π−x)+tan(x)−1=0
解答
x∈R无解
求解步骤
tan(2π−x)+tan(x)−1=0
使用三角恒等式改写
tan(2π−x)+tan(x)−1=0
使用三角恒等式改写
tan(2π−x)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=cos(2π−x)sin(2π−x)
使用角差恒等式: sin(s−t)=sin(s)cos(t)−cos(s)sin(t)=cos(2π−x)sin(2π)cos(x)−cos(2π)sin(x)
使用角差恒等式: cos(s−t)=cos(s)cos(t)+sin(s)sin(t)=cos(2π)cos(x)+sin(2π)sin(x)sin(2π)cos(x)−cos(2π)sin(x)
化简 cos(2π)cos(x)+sin(2π)sin(x)sin(2π)cos(x)−cos(2π)sin(x):sin(x)cos(x)
cos(2π)cos(x)+sin(2π)sin(x)sin(2π)cos(x)−cos(2π)sin(x)
sin(2π)cos(x)−cos(2π)sin(x)=cos(x)
sin(2π)cos(x)−cos(2π)sin(x)
sin(2π)cos(x)=cos(x)
sin(2π)cos(x)
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⋅cos(x)
乘以:1⋅cos(x)=cos(x)=cos(x)
cos(2π)sin(x)=0
cos(2π)sin(x)
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⋅sin(x)
使用法则 0⋅a=0=0
=cos(x)−0
cos(x)−0=cos(x)=cos(x)
=cos(2π)cos(x)+sin(2π)sin(x)cos(x)
cos(2π)cos(x)+sin(2π)sin(x)=sin(x)
cos(2π)cos(x)+sin(2π)sin(x)
cos(2π)cos(x)=0
cos(2π)cos(x)
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(2π)sin(x)=sin(x)
sin(2π)sin(x)
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:1⋅sin(x)=sin(x)=sin(x)
=0+sin(x)
0+sin(x)=sin(x)=sin(x)
=sin(x)cos(x)
=sin(x)cos(x)
sin(x)cos(x)+tan(x)−1=0
sin(x)cos(x)+tan(x)−1=0
化简 sin(x)cos(x)+tan(x)−1:sin(x)cos(x)+tan(x)sin(x)−sin(x)
sin(x)cos(x)+tan(x)−1
将项转换为分式: tan(x)=sin(x)tan(x)sin(x),1=sin(x)1sin(x)=sin(x)cos(x)+sin(x)tan(x)sin(x)−sin(x)1⋅sin(x)
因为分母相等,所以合并分式: ca±cb=ca±b=sin(x)cos(x)+tan(x)sin(x)−1⋅sin(x)
乘以:1⋅sin(x)=sin(x)=sin(x)cos(x)+tan(x)sin(x)−sin(x)
sin(x)cos(x)+tan(x)sin(x)−sin(x)=0
g(x)f(x)=0⇒f(x)=0cos(x)+tan(x)sin(x)−sin(x)=0
用 sin, cos 表示
cos(x)−sin(x)+sin(x)tan(x)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=cos(x)−sin(x)+sin(x)cos(x)sin(x)
化简 cos(x)−sin(x)+sin(x)cos(x)sin(x):cos(x)cos2(x)−sin(x)cos(x)+sin2(x)
cos(x)−sin(x)+sin(x)cos(x)sin(x)
sin(x)cos(x)sin(x)=cos(x)sin2(x)
sin(x)cos(x)sin(x)
分式相乘: a⋅cb=ca⋅b=cos(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)
=cos(x)sin2(x)
=cos(x)−sin(x)+cos(x)sin2(x)
将项转换为分式: cos(x)=cos(x)cos(x)cos(x),sin(x)=cos(x)sin(x)cos(x)=cos(x)cos(x)cos(x)−cos(x)sin(x)cos(x)+cos(x)sin2(x)
因为分母相等,所以合并分式: ca±cb=ca±b=cos(x)cos(x)cos(x)−sin(x)cos(x)+sin2(x)
cos(x)cos(x)−sin(x)cos(x)+sin2(x)=cos2(x)−sin(x)cos(x)+sin2(x)
cos(x)cos(x)−sin(x)cos(x)+sin2(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)−sin(x)cos(x)+sin2(x)
=cos(x)cos2(x)−sin(x)cos(x)+sin2(x)
=cos(x)cos2(x)−sin(x)cos(x)+sin2(x)
cos(x)cos2(x)+sin2(x)−cos(x)sin(x)=0
g(x)f(x)=0⇒f(x)=0cos2(x)+sin2(x)−cos(x)sin(x)=0
使用三角恒等式改写
−cos(x)sin(x)+1
使用倍角公式: 2sin(x)cos(x)=sin(2x)sin(x)cos(x)=2sin(2x)=1−2sin(2x)
1−2sin(2x)=0
将 1到右边
1−2sin(2x)=0
两边减去 11−2sin(2x)−1=0−1
化简−2sin(2x)=−1
−2sin(2x)=−1
在两边乘以 2
−2sin(2x)=−1
在两边乘以 22(−2sin(2x))=2(−1)
化简−sin(2x)=−2
−sin(2x)=−2
两边除以 −1
−sin(2x)=−2
两边除以 −1−1−sin(2x)=−1−2
化简sin(2x)=2
sin(2x)=2
−1≤sin(x)≤1x∈R无解