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受欢迎的三角函数 >

solvefor x,u=cos(x^2y)

解答

求解

x, u = cos (x^{2} y)

解答

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

求解步骤

u = cos (x^{2} y)

交换两边

cos (x^{2} y) = u

使用反三角函数性质

cos (x^{2} y) = u

cos (x^{2} y) = u

的通解

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

x^{2} y = arccos (u) + 2 πn, x^{2} y = - arccos (u) + 2 πn

x^{2} y = arccos (u) + 2 πn, x^{2} y = - arccos (u) + 2 πn

解

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

x^{2} y = arccos (u) + 2 πn

两边除以

y; y \neq = 0

x^{2} y = arccos (u) + 2 πn

两边除以

y; y \neq = 0

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

化简

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

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

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

化简

\frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{arccos ( u ) + 2 πn}{y}

\frac{arccos ( u )}{y} + \frac{2 πn}{y}

合并分式

\frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{arccos ( u ) + 2 πn}{y}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{arccos ( u ) + 2 πn}{y}

= \frac{arccos ( u ) + 2 πn}{y}

化简

- \frac{arccos ( u )}{y} + \frac{2 πn}{y} : - \frac{arccos ( u ) + 2 πn}{y}

- \frac{arccos ( u )}{y} + \frac{2 πn}{y}

合并分式

\frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{arccos ( u ) + 2 πn}{y}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{arccos ( u ) + 2 πn}{y}

= - \frac{2 πn + arccos ( u )}{y}

= - \frac{arccos ( u ) + 2 πn}{y}

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

解

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

x^{2} y = - arccos (u) + 2 πn

两边除以

y; y \neq = 0

x^{2} y = - arccos (u) + 2 πn

两边除以

y; y \neq = 0

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

化简

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

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

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

化简

- \frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{- arccos ( u ) + 2 πn}{y}

- \frac{arccos ( u )}{y} + \frac{2 πn}{y}

合并分式

- \frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{- arccos ( u ) + 2 πn}{y}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- arccos ( u ) + 2 πn}{y}

= \frac{- arccos ( u ) + 2 πn}{y}

化简

- - \frac{arccos ( u )}{y} + \frac{2 πn}{y} : - \frac{- arccos ( u ) + 2 πn}{y}

- - \frac{arccos ( u )}{y} + \frac{2 πn}{y}

合并分式

- \frac{arccos ( u )}{y} + \frac{2 πn}{y} : \frac{- arccos ( u ) + 2 πn}{y}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- arccos ( u ) + 2 πn}{y}

= - \frac{2 πn - arccos ( u )}{y}

= - \frac{- arccos ( u ) + 2 πn}{y}

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

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

作图

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