解答
求解 x,cos2(x)+cos2(y)cos2(z)=1
解答
x=arcsin(−cos(y)cos(z))+2πn,x=π+arcsin(cos(y)cos(z))+2πn,x=arcsin(cos(y)cos(z))+2πn,x=π+arcsin(−cos(y)cos(z))+2πn
求解步骤
cos2(x)+cos2(y)cos2(z)=1
两边减去 1cos2(x)+cos2(y)cos2(z)−1=0
使用三角恒等式改写
−1+cos2(x)+cos2(y)cos2(z)
使用毕达哥拉斯恒等式: 1=cos2(x)+sin2(x)1−cos2(x)=sin2(x)=cos2(y)cos2(z)−sin2(x)
−sin2(x)+cos2(y)cos2(z)=0
分解 −sin2(x)+cos2(y)cos2(z):(cos(y)cos(z)+sin(x))(cos(y)cos(z)−sin(x))
−sin2(x)+cos2(y)cos2(z)
将 cos2(y)cos2(z) 改写为 (cos(y)cos(z))2
cos2(y)cos2(z)
使用指数法则: ambm=(ab)mcos2(y)cos2(z)=(cos(y)cos(z))2=(cos(y)cos(z))2
=−sin2(x)+(cos(y)cos(z))2
使用平方差公式: x2−y2=(x+y)(x−y)(cos(y)cos(z))2−sin2(x)=(cos(y)cos(z)+sin(x))(cos(y)cos(z)−sin(x))=(cos(y)cos(z)+sin(x))(cos(y)cos(z)−sin(x))
(cos(y)cos(z)+sin(x))(cos(y)cos(z)−sin(x))=0
分别求解每个部分cos(y)cos(z)+sin(x)=0orcos(y)cos(z)−sin(x)=0
cos(y)cos(z)+sin(x)=0:x=arcsin(−cos(y)cos(z))+2πn,x=π+arcsin(cos(y)cos(z))+2πn
cos(y)cos(z)+sin(x)=0
将 cos(y)cos(z)到右边
cos(y)cos(z)+sin(x)=0
两边减去 cos(y)cos(z)cos(y)cos(z)+sin(x)−cos(y)cos(z)=0−cos(y)cos(z)
化简sin(x)=−cos(y)cos(z)
sin(x)=−cos(y)cos(z)
使用反三角函数性质
sin(x)=−cos(y)cos(z)
sin(x)=−cos(y)cos(z)的通解sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−cos(y)cos(z))+2πn,x=π+arcsin(cos(y)cos(z))+2πn
x=arcsin(−cos(y)cos(z))+2πn,x=π+arcsin(cos(y)cos(z))+2πn
cos(y)cos(z)−sin(x)=0:x=arcsin(cos(y)cos(z))+2πn,x=π+arcsin(−cos(y)cos(z))+2πn
cos(y)cos(z)−sin(x)=0
将 cos(y)cos(z)到右边
cos(y)cos(z)−sin(x)=0
两边减去 cos(y)cos(z)cos(y)cos(z)−sin(x)−cos(y)cos(z)=0−cos(y)cos(z)
化简−sin(x)=−cos(y)cos(z)
−sin(x)=−cos(y)cos(z)
两边除以 −1
−sin(x)=−cos(y)cos(z)
两边除以 −1−1−sin(x)=−1−cos(y)cos(z)
化简sin(x)=cos(y)cos(z)
sin(x)=cos(y)cos(z)
使用反三角函数性质
sin(x)=cos(y)cos(z)
sin(x)=cos(y)cos(z)的通解sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin(cos(y)cos(z))+2πn,x=π+arcsin(−cos(y)cos(z))+2πn
x=arcsin(cos(y)cos(z))+2πn,x=π+arcsin(−cos(y)cos(z))+2πn
合并所有解x=arcsin(−cos(y)cos(z))+2πn,x=π+arcsin(cos(y)cos(z))+2πn,x=arcsin(cos(y)cos(z))+2πn,x=π+arcsin(−cos(y)cos(z))+2πn