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受欢迎的三角函数 >

solvefor x,z=cos(xy)

解答

求解

x, z = cos (x y)

解答

x = \frac{arccos ( z )}{y} + \frac{2 πn}{y}, x = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}

求解步骤

z = cos (x y)

交换两边

cos (x y) = z

使用反三角函数性质

cos (x y) = z

cos (x y) = z

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x y = arccos (z) + 2 πn, x y = - arccos (z) + 2 πn

x y = arccos (z) + 2 πn, x y = - arccos (z) + 2 πn

解

x y = arccos (z) + 2 πn : x = \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

x y = arccos (z) + 2 πn

两边除以

y; y \neq = 0

x y = arccos (z) + 2 πn

两边除以

y; y \neq = 0

\frac{x y}{y} = \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

化简

x = \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

x = \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

解

x y = - arccos (z) + 2 πn : x = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

x y = - arccos (z) + 2 πn

两边除以

y; y \neq = 0

x y = - arccos (z) + 2 πn

两边除以

y; y \neq = 0

\frac{x y}{y} = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

化简

x = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

x = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}; y \neq = 0

x = \frac{arccos ( z )}{y} + \frac{2 πn}{y}, x = - \frac{arccos ( z )}{y} + \frac{2 πn}{y}

作图

流行的例子

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