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受欢迎的三角函数 >

solvefor x,sqrt(1+sin^3(xy^2))=y

解答

求解

x, 1 + sin^{3} (x y^{2}) = y

解答

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}

+1

弧度

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 π}{y ^{2}} n, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 π}{y ^{2}} n

求解步骤

1 + sin^{3} (x y^{2}) = y

两边进行平方:

1 + sin^{3} (x y^{2}) = y^{2}

1 + sin^{3} (x y^{2}) = y

(1 + sin^{3} (x y^{2}))^{2} = y^{2}

展开

(1 + sin^{3} (x y^{2}))^{2} : 1 + sin^{3} (x y^{2})

(1 + sin^{3} (x y^{2}))^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= ((1 + sin^{3} (x y^{2}))^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (1 + sin^{3} (x y^{2}))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 1 + sin^{3} (x y^{2})

1 + sin^{3} (x y^{2}) = y^{2}

1 + sin^{3} (x y^{2}) = y^{2}

解

1 + sin^{3} (x y^{2}) = y^{2} : sin (x y^{2}) = 3 y^{2} - 1

1 + sin^{3} (x y^{2}) = y^{2}

将

1

到右边

1 + sin^{3} (x y^{2}) = y^{2}

两边减去

1

1 + sin^{3} (x y^{2}) - 1 = y^{2} - 1

化简

sin^{3} (x y^{2}) = y^{2} - 1

sin^{3} (x y^{2}) = y^{2} - 1

对于

x^{n} = f (a)

，n 为奇数，解为

x = n f (a)

sin (x y^{2}) = 3 y^{2} - 1

sin (x y^{2}) = 3 y^{2} - 1

验证解:

sin (x y^{2}) = 3 y^{2} - 1 {y \geq 0}

将它们代入

1 + sin^{3} (x y^{2}) = y

检验解是否符合

去除与方程不符的解。

代入

sin (x y^{2}) = 3 y^{2} - 1 : 1 + (3 y^{2} - 1)^{3} = y \Rightarrow y \geq 0

1 + (3 y^{2} - 1)^{3} = y

两边进行平方:

y^{2} = y^{2}

1 + (3 y^{2} - 1)^{3} = y

(1 + (3 y^{2} - 1)^{3})^{2} = y^{2}

展开

(1 + (3 y^{2} - 1)^{3})^{2} : y^{2}

(1 + (3 y^{2} - 1)^{3})^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= ((1 + (3 y^{2} - 1)^{3})^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (1 + (3 y^{2} - 1)^{3})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 1 + (3 y^{2} - 1)^{3}

展开

1 + (3 y^{2} - 1)^{3} : y^{2}

1 + (3 y^{2} - 1)^{3}

(3 y^{2} - 1)^{3} = y^{2} - 1

(3 y^{2} - 1)^{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

分式相乘:

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

= \frac{1 \cdot 3}{3}

约分:

3

= 1

= y^{2} - 1

= 1 + y^{2} - 1

对同类项分组

= y^{2} + 1 - 1

1 - 1 = 0

= y^{2}

= y^{2}

y^{2} = y^{2}

y^{2} = y^{2}

两侧相等

对所有 y 为真

验证解:

y < 0

假

, y = 0

真

, y > 0

真

1 + (3 y^{2} - 1)^{3} = y

将定义域区间与解区间合并:

对所有 y 为真

找到函数区间:

y < 0, y = 0, y > 0

1 + (3 y^{2} - 1)^{3} = y

找到偶数根参量零值:

解

1 + 3 y^{2} - 1^{3} = 0 : y = 0

1 + (3 y^{2} - 1)^{3} = 0

因式分解

1 + (3 y^{2} - 1)^{3} : (3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1)

1 + (3 y^{2} - 1)^{3}

将

1

改写为

1^{3}

= (3 y^{2} - 1)^{3} + 1^{3}

使用立方和公式:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(3 y^{2} - 1)^{3} + 1^{3} = (3 y^{2} - 1 + 1) ((3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1)

= (3 y^{2} - 1 + 1) ((3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1)

(3 y^{2} - 1)^{2} = (y^{2} - 1)^{\frac{2}{3}}

(3 y^{2} - 1)^{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 2}

\frac{1}{3} \cdot 2 = \frac{2}{3}

\frac{1}{3} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{3}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{3}

= (y^{2} - 1)^{\frac{2}{3}}

= (3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1)

(3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

3 y^{2} - 1 + 1 = 0 or (y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0

解

3 y^{2} - 1 + 1 = 0 : y = 0

3 y^{2} - 1 + 1 = 0

将

1

到右边

3 y^{2} - 1 + 1 = 0

两边减去

1

3 y^{2} - 1 + 1 - 1 = 0 - 1

化简

3 y^{2} - 1 = - 1

3 y^{2} - 1 = - 1

对方程式两边

3

次方:

y^{2} - 1 = - 1

3 y^{2} - 1 = - 1

(3 y^{2} - 1)^{3} = (- 1)^{3}

展开

(3 y^{2} - 1)^{3} : y^{2} - 1

(3 y^{2} - 1)^{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

分式相乘:

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

= \frac{1 \cdot 3}{3}

约分:

3

= 1

= y^{2} - 1

展开

(- 1)^{3} : - 1

(- 1)^{3}

使用指数法则:

(- a)^{n} = - a^{n},

若

n

是奇数

(- 1)^{3} = - 1^{3}

= - 1^{3}

使用法则

1^{a} = 1

= - 1

y^{2} - 1 = - 1

y^{2} - 1 = - 1

解

y^{2} - 1 = - 1 : y = 0

y^{2} - 1 = - 1

将

1

到右边

y^{2} - 1 = - 1

两边加上

1

y^{2} - 1 + 1 = - 1 + 1

化简

y^{2} = 0

y^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

y = 0

y = 0

验证解:

y = 0

真

将它们代入

3 y^{2} - 1 + 1 = 0

检验解是否符合

去除与方程不符的解。

代入

y = 0 :

真

3 0^{2} - 1 + 1 = 0

3 0^{2} - 1 + 1 = 0

3 0^{2} - 1 + 1

使用法则

0^{a} = 0

0^{2} = 0

= 3 0 - 1 + 1

3 0 - 1 = - 1

3 0 - 1

数字相减:

0 - 1 = - 1

= 3 - 1

n - 1 = - 1,

若

n

是奇数

3 - 1 = - 1

= - 1

= - 1 + 1

数字相加/相减:

- 1 + 1 = 0

= 0

0 = 0

真

解是

y = 0

解

(y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0 : y \in R

无解

(y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0

使用以下的指数特性

: a^{\frac{n}{m}} = (m a)^{n}

(y^{2} - 1)^{\frac{2}{3}} = (3 y^{2} - 1)^{2}

(3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1 = 0

用

3 y^{2} - 1 = u

改写方程式

u^{2} - u + 1 = 0

解

u^{2} - u + 1 = 0 : u \in R

无解

u^{2} - u + 1 = 0

判别式

u^{2} - u + 1 = 0 : - 3

u^{2} - u + 1 = 0

对于

a x^{2} + b x + c = 0

形式的二次方程，根判别式为

b^{2} - 4 a c

若

a = 1, b = - 1, c = 1 : (- 1)^{2} - 4 \cdot 1 \cdot 1

(- 1)^{2} - 4 \cdot 1 \cdot 1

展开

(- 1)^{2} - 4 \cdot 1 \cdot 1 : - 3

(- 1)^{2} - 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

- 3

判别式在

u

内不能为负

\in R

解是

u \in R 无解

y \in R 无解

y = 0

验证解:

y = 0

真

将它们代入

1 + 3 y^{2} - 1^{3} = 0

检验解是否符合

去除与方程不符的解。

代入

y = 0 :

真

1 + (3 0^{2} - 1)^{3} = 0

1 + (3 0^{2} - 1)^{3} = 0

1 + (3 0^{2} - 1)^{3}

使用法则

0^{a} = 0

0^{2} = 0

= 1 + (3 0 - 1)^{3}

(3 0 - 1)^{3} = - 1

(3 0 - 1)^{3}

3 0 - 1 = - 1

3 0 - 1

数字相减:

0 - 1 = - 1

= 3 - 1

n - 1 = - 1,

若

n

是奇数

3 - 1 = - 1

= - 1

= (- 1)^{3}

使用指数法则:

(- a)^{n} = - a^{n},

若

n

是奇数

(- 1)^{3} = - 1^{3}

= - 1^{3}

使用法则

1^{a} = 1

= - 1

= 1 - 1

数字相减:

1 - 1 = 0

= 0

0 = 0

真

解是

y = 0

y = 0

区间围绕零点来定义:

y < 0, y = 0, y > 0

将区间与定义域合并

y < 0, y = 0, y > 0

将它们代入

1 + 3 y^{2} - 1^{3} = y

检验解是否符合

去除与方程不符的解。

代入

y < 0 : 1 + 3 y^{2} - 1^{3} \neq = y \Rightarrow

假

解是

y \geq 0

解是

sin (x y^{2}) = 3 y^{2} - 1 {y \geq 0}

使用反三角函数性质

sin (x y^{2}) = 3 y^{2} - 1

sin (x y^{2}) = 3 y^{2} - 1

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn, x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn, x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

解

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn : x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn

两边除以

y^{2}; y \neq = 0

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn

两边除以

y^{2}; y \neq = 0

\frac{x y ^{2}}{y ^{2}} = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

化简

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

解

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn : x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

两边除以

y^{2}; y \neq = 0

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

两边除以

y^{2}; y \neq = 0

\frac{x y ^{2}}{y ^{2}} = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

化简

x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}

作图

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