解答
求解 x,sin(x)+cos(y)=sin(x)cos(y)
解答
x=arccsc(cos(y)cos(y)−1)+2πn,x=π+arccsc(−cos(y)cos(y)−1)+2πn
求解步骤
sin(x)+cos(y)=sin(x)cos(y)
两边减去 sin(x)cos(y)sin(x)+cos(y)−cos(y)sin(x)=0
使用三角恒等式改写
cos(y)+sin(x)−cos(y)sin(x)
使用基本三角恒等式: sin(x)=csc(x)1=cos(y)+csc(x)1−cos(y)csc(x)1
化简 cos(y)+csc(x)1−cos(y)csc(x)1:cos(y)+csc(x)1−cos(y)
cos(y)+csc(x)1−cos(y)csc(x)1
cos(y)csc(x)1=csc(x)cos(y)
cos(y)csc(x)1
分式相乘: a⋅cb=ca⋅b=csc(x)1⋅cos(y)
乘以:1⋅cos(y)=cos(y)=csc(x)cos(y)
=cos(y)+csc(x)1−csc(x)cos(y)
合并分式 csc(x)1−csc(x)cos(y):csc(x)1−cos(y)
使用法则 ca±cb=ca±b=csc(x)1−cos(y)
=cos(y)+csc(x)−cos(y)+1
=cos(y)+csc(x)1−cos(y)
cos(y)+csc(x)1−cos(y)=0
在两边乘以 csc(x)
cos(y)+csc(x)1−cos(y)=0
在两边乘以 csc(x)cos(y)csc(x)+csc(x)1−cos(y)csc(x)=0⋅csc(x)
化简
cos(y)csc(x)+csc(x)1−cos(y)csc(x)=0⋅csc(x)
化简 csc(x)1−cos(y)csc(x):1−cos(y)
csc(x)1−cos(y)csc(x)
分式相乘: a⋅cb=ca⋅b=csc(x)(1−cos(y))csc(x)
约分:csc(x)=1−cos(y)
化简 0⋅csc(x):0
0⋅csc(x)
使用法则 0⋅a=0=0
cos(y)csc(x)+1−cos(y)=0
cos(y)csc(x)+1−cos(y)=0
cos(y)csc(x)+1−cos(y)=0
将 cos(y)到右边
cos(y)csc(x)+1−cos(y)=0
两边加上 cos(y)cos(y)csc(x)+1−cos(y)+cos(y)=0+cos(y)
化简cos(y)csc(x)+1=cos(y)
cos(y)csc(x)+1=cos(y)
将 1到右边
cos(y)csc(x)+1=cos(y)
两边减去 1cos(y)csc(x)+1−1=cos(y)−1
化简cos(y)csc(x)=cos(y)−1
cos(y)csc(x)=cos(y)−1
两边除以 cos(y);y=2π+2πn,y=23π+2πn
cos(y)csc(x)=cos(y)−1
两边除以 cos(y);y=2π+2πn,y=23π+2πncos(y)cos(y)csc(x)=cos(y)cos(y)−cos(y)1;y=2π+2πn,y=23π+2πn
化简
cos(y)cos(y)csc(x)=cos(y)cos(y)−cos(y)1
化简 cos(y)cos(y)csc(x):csc(x)
cos(y)cos(y)csc(x)
约分:cos(y)=csc(x)
化简 cos(y)cos(y)−cos(y)1:cos(y)cos(y)−1
cos(y)cos(y)−cos(y)1
使用法则 ca±cb=ca±b=cos(y)cos(y)−1
csc(x)=cos(y)cos(y)−1;y=2π+2πn,y=23π+2πn
csc(x)=cos(y)cos(y)−1;y=2π+2πn,y=23π+2πn
csc(x)=cos(y)cos(y)−1;y=2π+2πn,y=23π+2πn
使用反三角函数性质
csc(x)=cos(y)cos(y)−1
csc(x)=cos(y)cos(y)−1的通解csc(x)=a⇒x=arccsc(a)+2πn,x=π+arccsc(a)+2πnx=arccsc(cos(y)cos(y)−1)+2πn,x=π+arccsc(−cos(y)cos(y)−1)+2πn
x=arccsc(cos(y)cos(y)−1)+2πn,x=π+arccsc(−cos(y)cos(y)−1)+2πn