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受欢迎的三角函数 >

solvefor x,e^{4y}=cos(4x+y)

解答

求解

x, e^{4 y} = cos (4 x + y)

解答

x = \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}, x = - \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

求解步骤

e^{4 y} = cos (4 x + y)

交换两边

cos (4 x + y) = e^{4 y}

使用反三角函数性质

cos (4 x + y) = e^{4 y}

cos (4 x + y) = e^{4 y}

的通解

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

4 x + y = arccos (e^{4 y}) + 2 πn, 4 x + y = - arccos (e^{4 y}) + 2 πn

4 x + y = arccos (e^{4 y}) + 2 πn, 4 x + y = - arccos (e^{4 y}) + 2 πn

解

4 x + y = arccos (e^{4 y}) + 2 πn : x = \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

4 x + y = arccos (e^{4 y}) + 2 πn

将

y

到右边

4 x + y = arccos (e^{4 y}) + 2 πn

两边减去

y

4 x + y - y = arccos (e^{4 y}) + 2 πn - y

化简

4 x = arccos (e^{4 y}) + 2 πn - y

4 x = arccos (e^{4 y}) + 2 πn - y

两边除以

4

4 x = arccos (e^{4 y}) + 2 πn - y

两边除以

4

\frac{4 x}{4} = \frac{arccos ( e ^{4 y} )}{4} + \frac{2 πn}{4} - \frac{y}{4}

化简

x = \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

x = \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

解

4 x + y = - arccos (e^{4 y}) + 2 πn : x = - \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

4 x + y = - arccos (e^{4 y}) + 2 πn

将

y

到右边

4 x + y = - arccos (e^{4 y}) + 2 πn

两边减去

y

4 x + y - y = - arccos (e^{4 y}) + 2 πn - y

化简

4 x = - arccos (e^{4 y}) + 2 πn - y

4 x = - arccos (e^{4 y}) + 2 πn - y

两边除以

4

4 x = - arccos (e^{4 y}) + 2 πn - y

两边除以

4

\frac{4 x}{4} = - \frac{arccos ( e ^{4 y} )}{4} + \frac{2 πn}{4} - \frac{y}{4}

化简

x = - \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

x = - \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

x = \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}, x = - \frac{arccos ( e ^{4 y} )}{4} + \frac{πn}{2} - \frac{y}{4}

作图

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