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受欢迎的三角函数 >

cos^6(x)=-cos^2(x)

解答

cos^{6} (x) = - cos^{2} (x)

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

cos^{6} (x) = - cos^{2} (x)

用替代法求解

cos^{6} (x) = - cos^{2} (x)

令

: cos (x) = u

u^{6} = - u^{2}

u^{6} = - u^{2} : u = 0, u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i, u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

u^{6} = - u^{2}

将

u^{2}

para o lado esquerdo

u^{6} = - u^{2}

两边加上

u^{2}

u^{6} + u^{2} = - u^{2} + u^{2}

化简

u^{6} + u^{2} = 0

u^{6} + u^{2} = 0

用

a = u^{2}

和

a^{3} = u^{6}

改写方程式

a^{3} + a = 0

解

a^{3} + a = 0 : a = 0, a = i, a = - i

a^{3} + a = 0

因式分解

a^{3} + a : a (a^{2} + 1)

a^{3} + a

使用指数法则:

a^{b + c} = a^{b} a^{c}

a^{3} = a^{2} a

= a^{2} a + a

因式分解出通项

a

= a (a^{2} + 1)

a (a^{2} + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

a = 0 or a^{2} + 1 = 0

解

a^{2} + 1 = 0 : a = i, a = - i

a^{2} + 1 = 0

将

1

到右边

a^{2} + 1 = 0

两边减去

1

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

化简

a^{2} = - 1

a^{2} = - 1

对于

x^{2} = f (a)

解为

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

a = - 1, a = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

a = i, a = - i

解为

a = 0, a = i, a = - i

a = 0, a = i, a = - i

代回

a = u^{2}

，求解

u

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

解

u^{2} = i : u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i

u^{2} = i

替代

u = a + bi

(a + bi)^{2} = i

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = i

将

i

改写成标准复数形式:

0 + i

(a^{2} - b^{2}) + 2 iab = 0 + i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = 0 2 ab = 1]

[a^{2} - b^{2} = 0 2 ab = 1] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = 0 2 ab = 1]

对于

2 ab = 1

将

a

移到一边:

a = \frac{1}{2 b}

2 ab = 1

两边除以

2 b

2 ab = 1

两边除以

2 b

\frac{2 ab}{2 b} = \frac{1}{2 b}

化简

a = \frac{1}{2 b}

a = \frac{1}{2 b}

将解

a = \frac{1}{2 b}

代入

a^{2} - b^{2} = 0

对于

a^{2} - b^{2} = 0

，用

\frac{1}{2 b}

替代

a : b = \frac{1}{2}, b = - \frac{1}{2}

对于

a^{2} - b^{2} = 0

，用

\frac{1}{2 b}

替代

a

(\frac{1}{2 b})^{2} - b^{2} = 0

解

(\frac{1}{2 b})^{2} - b^{2} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

(\frac{1}{2 b})^{2} - b^{2} = 0

化简

(\frac{1}{2 b})^{2} : \frac{1}{4 b ^{2}}

(\frac{1}{2 b})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{( 2 b ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(2 b)^{2} = 2^{2} b^{2}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4 b ^{2}}

\frac{1}{4 b ^{2}} - b^{2} = 0

在两边乘以

4 b^{2}

\frac{1}{4 b ^{2}} - b^{2} = 0

在两边乘以

4 b^{2}

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

化简

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

化简

\frac{1}{4 b ^{2}} \cdot 4 b^{2} : 1

\frac{1}{4 b ^{2}} \cdot 4 b^{2}

分式相乘:

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

= \frac{1 \cdot 4 b ^{2}}{4 b ^{2}}

约分:

4

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

约分:

b^{2}

= 1

化简

- b^{2} \cdot 4 b^{2} : - 4 b^{4}

- b^{2} \cdot 4 b^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 4 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 4 b^{4}

化简

0 \cdot 4 b^{2} : 0

0 \cdot 4 b^{2}

使用法则

0 \cdot a = 0

= 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

解

1 - 4 b^{4} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

1 - 4 b^{4} = 0

将

1

到右边

1 - 4 b^{4} = 0

两边减去

1

1 - 4 b^{4} - 1 = 0 - 1

化简

- 4 b^{4} = - 1

- 4 b^{4} = - 1

两边除以

- 4

- 4 b^{4} = - 1

两边除以

- 4

\frac{- 4 b ^{4}}{- 4} = \frac{- 1}{- 4}

化简

b^{4} = \frac{1}{4}

b^{4} = \frac{1}{4}

对于

x^{n} = f (a)

，n 为偶数，解为

使用根式运算法则:

使用根式运算法则:

因式分解数字:

4 = 2^{2}

使用指数法则:

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

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

使用根式运算法则:

假定

a \geq 0

= 2

= \frac{1}{2}

使用根式运算法则:

使用根式运算法则:

因式分解数字:

4 = 2^{2}

使用指数法则:

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

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

使用根式运算法则:

假定

a \geq 0

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(\frac{1}{2 b})^{2} - b^{2}

的分母，令其等于零

解

2 b = 0 : b = 0

2 b = 0

两边除以

2

2 b = 0

两边除以

2

\frac{2 b}{2} = \frac{0}{2}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{1}{2}, b = - \frac{1}{2}

将解

b = \frac{1}{2}, b = - \frac{1}{2}

代入

2 ab = 1

对于

2 ab = 1

，用

\frac{1}{2}

替代

b : a = \frac{1}{2}

对于

2 ab = 1

，用

\frac{1}{2}

替代

b

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

解

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

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

在两边乘以

2

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

在两边乘以

2

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

化简

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

化简

2 a \frac{1}{2} 2 : 2 a

2 a \frac{1}{2} 2

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

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

交叉约去公约数:

2

= 2 a \cdot 1

使用法则:

a \cdot 1 = a

= 2 a

化简

1 \cdot 2 : 2

1 \cdot 2

使用法则:

1 \cdot a = a

= 2

2 a = 2

2 a = 2

2 a = 2

两边除以

2

2 a = 2

两边除以

2

\frac{2 a}{2} = \frac{2}{2}

化简

\frac{2 a}{2} = \frac{2}{2}

化简

\frac{2 a}{2} : a

\frac{2 a}{2}

约分:

2

= a

化简

\frac{2}{2} : \frac{1}{2}

\frac{2}{2}

使用根式运算法则:

a = a a

2 = 22

= \frac{2}{2 2}

约分:

2

= \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

对于

2 ab = 1

，用

- \frac{1}{2}

替代

b : a = - \frac{1}{2}

对于

2 ab = 1

，用

- \frac{1}{2}

替代

b

2 a (- \frac{1}{2}) = 1

解

2 a (- \frac{1}{2}) = 1 : a = - \frac{1}{2}

2 a (- \frac{1}{2}) = 1

两边除以

2 (- \frac{1}{2})

2 a (- \frac{1}{2}) = 1

两边除以

2 (- \frac{1}{2})

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{1}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{1}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 a (- \frac{1}{2}) = - 2 a \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

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

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

约分:

- 2

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

约分:

\frac{1}{2}

= a

化简

\frac{1}{2 ( - \frac{1}{2} )} : - \frac{1}{2}

\frac{1}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

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

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

- 2 \cdot \frac{1}{2}

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

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

使用分式法则:

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

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

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

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

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

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

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

数字相乘:

2 \cdot 1 = 2

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

使用法则:

1 \cdot a = a

1 \cdot 2 = 2

= \frac{2}{2}

= \frac{2}{2}

使用根式运算法则:

a = a a

2 = 22

= \frac{2 2}{2}

约分:

2

= 2

= - 2

= \frac{1}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

将解代入原方程进行验证

将它们代入

a^{2} - b^{2} = 0

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{1}{2}, b = - \frac{1}{2}

的解:

真

a^{2} - b^{2} = 0

代入

a = - \frac{1}{2}, b = - \frac{1}{2}

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

整理后得

0 = 0

真

检验

a = \frac{1}{2}, b = \frac{1}{2}

的解:

真

a^{2} - b^{2} = 0

代入

a = \frac{1}{2}, b = \frac{1}{2}

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

整理后得

0 = 0

真

将它们代入

2 ab = 1

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{1}{2}, b = - \frac{1}{2}

的解:

真

2 ab = 1

代入

a = - \frac{1}{2}, b = - \frac{1}{2}

2 (- \frac{1}{2}) (- \frac{1}{2}) = 1

整理后得

1 = 1

真

检验

a = \frac{1}{2}, b = \frac{1}{2}

的解:

真

2 ab = 1

代入

a = \frac{1}{2}, b = \frac{1}{2}

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

整理后得

1 = 1

真

因而，

a^{2} - b^{2} = 0, 2 ab = 1

最后的解是

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

u = a + bi

代回

u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i

解

u^{2} = - i : u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

u^{2} = - i

替代

u = a + bi

(a + bi)^{2} = - i

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = - i

将

- i

改写成标准复数形式:

0 - i

(a^{2} - b^{2}) + 2 iab = 0 - i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = 0 2 ab = - 1]

[a^{2} - b^{2} = 0 2 ab = - 1] : (a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = 0 2 ab = - 1]

对于

2 ab = - 1

将

a

移到一边:

a = - \frac{1}{2 b}

2 ab = - 1

两边除以

2 b

2 ab = - 1

两边除以

2 b

\frac{2 ab}{2 b} = \frac{- 1}{2 b}

化简

a = - \frac{1}{2 b}

a = - \frac{1}{2 b}

将解

a = - \frac{1}{2 b}

代入

a^{2} - b^{2} = 0

对于

a^{2} - b^{2} = 0

，用

- \frac{1}{2 b}

替代

a : b = \frac{1}{2}, b = - \frac{1}{2}

对于

a^{2} - b^{2} = 0

，用

- \frac{1}{2 b}

替代

a

(- \frac{1}{2 b})^{2} - b^{2} = 0

解

(- \frac{1}{2 b})^{2} - b^{2} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

(- \frac{1}{2 b})^{2} - b^{2} = 0

化简

(- \frac{1}{2 b})^{2} : \frac{1}{4 b ^{2}}

(- \frac{1}{2 b})^{2}

使用指数法则:

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

若

n

是偶数

(- \frac{1}{2 b})^{2} = (\frac{1}{2 b})^{2}

= (\frac{1}{2 b})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{( 2 b ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(2 b)^{2} = 2^{2} b^{2}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4 b ^{2}}

\frac{1}{4 b ^{2}} - b^{2} = 0

在两边乘以

4 b^{2}

\frac{1}{4 b ^{2}} - b^{2} = 0

在两边乘以

4 b^{2}

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

化简

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

化简

\frac{1}{4 b ^{2}} \cdot 4 b^{2} : 1

\frac{1}{4 b ^{2}} \cdot 4 b^{2}

分式相乘:

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

= \frac{1 \cdot 4 b ^{2}}{4 b ^{2}}

约分:

4

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

约分:

b^{2}

= 1

化简

- b^{2} \cdot 4 b^{2} : - 4 b^{4}

- b^{2} \cdot 4 b^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 4 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 4 b^{4}

化简

0 \cdot 4 b^{2} : 0

0 \cdot 4 b^{2}

使用法则

0 \cdot a = 0

= 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

解

1 - 4 b^{4} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

1 - 4 b^{4} = 0

将

1

到右边

1 - 4 b^{4} = 0

两边减去

1

1 - 4 b^{4} - 1 = 0 - 1

化简

- 4 b^{4} = - 1

- 4 b^{4} = - 1

两边除以

- 4

- 4 b^{4} = - 1

两边除以

- 4

\frac{- 4 b ^{4}}{- 4} = \frac{- 1}{- 4}

化简

b^{4} = \frac{1}{4}

b^{4} = \frac{1}{4}

对于

x^{n} = f (a)

，n 为偶数，解为

使用根式运算法则:

使用根式运算法则:

因式分解数字:

4 = 2^{2}

使用指数法则:

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

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

使用根式运算法则:

假定

a \geq 0

= 2

= \frac{1}{2}

使用根式运算法则:

使用根式运算法则:

因式分解数字:

4 = 2^{2}

使用指数法则:

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

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

使用根式运算法则:

假定

a \geq 0

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(- \frac{1}{2 b})^{2} - b^{2}

的分母，令其等于零

解

2 b = 0 : b = 0

2 b = 0

两边除以

2

2 b = 0

两边除以

2

\frac{2 b}{2} = \frac{0}{2}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{1}{2}, b = - \frac{1}{2}

将解

b = \frac{1}{2}, b = - \frac{1}{2}

代入

2 ab = - 1

对于

2 ab = - 1

，用

\frac{1}{2}

替代

b : a = - \frac{1}{2}

对于

2 ab = - 1

，用

\frac{1}{2}

替代

b

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

解

2 a \frac{1}{2} = - 1 : a = - \frac{1}{2}

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

在两边乘以

2

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

在两边乘以

2

2 a \frac{1}{2} 2 = (- 1) 2

化简

2 a \frac{1}{2} 2 = (- 1) 2

化简

2 a \frac{1}{2} 2 : 2 a

2 a \frac{1}{2} 2

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

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

交叉约去公约数:

2

= 2 a \cdot 1

使用法则:

a \cdot 1 = a

= 2 a

化简

(- 1) 2 : - 2

(- 1) 2

使用法则:

(- a) = - a

(- 1) = - 1

= - 1 \cdot 2

使用法则:

1 \cdot a = a

= - 2

2 a = - 2

2 a = - 2

2 a = - 2

两边除以

2

2 a = - 2

两边除以

2

\frac{2 a}{2} = \frac{- 2}{2}

化简

\frac{2 a}{2} = \frac{- 2}{2}

化简

\frac{2 a}{2} : a

\frac{2 a}{2}

约分:

2

= a

化简

\frac{- 2}{2} : - \frac{1}{2}

\frac{- 2}{2}

使用根式运算法则:

a = a a

2 = 22

= \frac{- 2}{2 2}

约分:

2

= \frac{- 1}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

对于

2 ab = - 1

，用

- \frac{1}{2}

替代

b : a = \frac{1}{2}

对于

2 ab = - 1

，用

- \frac{1}{2}

替代

b

2 a (- \frac{1}{2}) = - 1

解

2 a (- \frac{1}{2}) = - 1 : a = \frac{1}{2}

2 a (- \frac{1}{2}) = - 1

两边除以

2 (- \frac{1}{2})

2 a (- \frac{1}{2}) = - 1

两边除以

2 (- \frac{1}{2})

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- 1}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- 1}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 a (- \frac{1}{2}) = - 2 a \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

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

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

约分:

- 2

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

约分:

\frac{1}{2}

= a

化简

\frac{- 1}{2 ( - \frac{1}{2} )} : \frac{1}{2}

\frac{- 1}{2 ( - \frac{1}{2} )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2 ( - \frac{1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

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

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

- 2 \cdot \frac{1}{2}

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

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

使用分式法则:

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

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

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

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

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

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

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

数字相乘:

2 \cdot 1 = 2

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

使用法则:

1 \cdot a = a

1 \cdot 2 = 2

= \frac{2}{2}

= \frac{2}{2}

使用根式运算法则:

a = a a

2 = 22

= \frac{2 2}{2}

约分:

2

= 2

= - 2

= - \frac{1}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

\frac{1}{- 2} = - \frac{1}{2}

= - (- \frac{1}{2})

使用法则:

- (- a) = a

- (- \frac{1}{2}) = \frac{1}{2}

= \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

将解代入原方程进行验证

将它们代入

a^{2} - b^{2} = 0

检验解是否符合

去除与方程不符的解。

检验

a = \frac{1}{2}, b = - \frac{1}{2}

的解:

真

a^{2} - b^{2} = 0

代入

a = \frac{1}{2}, b = - \frac{1}{2}

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

整理后得

0 = 0

真

检验

a = - \frac{1}{2}, b = \frac{1}{2}

的解:

真

a^{2} - b^{2} = 0

代入

a = - \frac{1}{2}, b = \frac{1}{2}

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

整理后得

0 = 0

真

将它们代入

2 ab = - 1

检验解是否符合

去除与方程不符的解。

检验

a = \frac{1}{2}, b = - \frac{1}{2}

的解:

真

2 ab = - 1

代入

a = \frac{1}{2}, b = - \frac{1}{2}

2 \cdot \frac{1}{2} (- \frac{1}{2}) = - 1

整理后得

- 1 = - 1

真

检验

a = - \frac{1}{2}, b = \frac{1}{2}

的解:

真

2 ab = - 1

代入

a = - \frac{1}{2}, b = \frac{1}{2}

2 (- \frac{1}{2}) \frac{1}{2} = - 1

整理后得

- 1 = - 1

真

因而，

a^{2} - b^{2} = 0, 2 ab = - 1

最后的解是

(a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

u = a + bi

代回

u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

解为

u = 0, u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i, u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{1}{2} i, cos (x) = - \frac{1}{2} - \frac{1}{2} i, cos (x) = - \frac{1}{2} + \frac{1}{2} i, cos (x) = \frac{1}{2} - \frac{1}{2} i

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{1}{2} i, cos (x) = - \frac{1}{2} - \frac{1}{2} i, cos (x) = - \frac{1}{2} + \frac{1}{2} i, cos (x) = \frac{1}{2} - \frac{1}{2} i

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = \frac{1}{2} + \frac{1}{2} i :

无解

cos (x) = \frac{1}{2} + \frac{1}{2} i

化简

\frac{1}{2} + \frac{1}{2} i : \frac{2}{2} + i \frac{2}{2}

\frac{1}{2} + \frac{1}{2} i

乘

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

因为分母相等，所以合并分式:

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

= \frac{1 + i}{2}

\frac{1 + i}{2}

有理化:

\frac{2 ( 1 + i )}{2}

\frac{1 + i}{2}

乘以共轭根式

\frac{2}{2}

= \frac{( 1 + i ) 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2 ( 1 + i )}{2}

= \frac{2 ( 1 + i )}{2}

将

\frac{2 ( 1 + i )}{2}

改写成标准复数形式:

\frac{2}{2} + \frac{2}{2} i

\frac{2 ( 1 + i )}{2}

使用根式运算法则:

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

= \frac{2 ^{\frac{1}{2}} ( 1 + i )}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{1 + i}{2 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1 + i}{2 ^{\frac{1}{2}}}

使用根式运算法则:

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

= \frac{1 + i}{2}

使用分式法则:

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

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

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

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{1}{2} + \frac{2}{2} i

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

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2} + \frac{2}{2} i

= \frac{2}{2} + \frac{2}{2} i

无解

cos (x) = - \frac{1}{2} - \frac{1}{2} i :

无解

cos (x) = - \frac{1}{2} - \frac{1}{2} i

化简

- \frac{1}{2} - \frac{1}{2} i : - \frac{2}{2} - i \frac{2}{2}

- \frac{1}{2} - \frac{1}{2} i

乘

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= - \frac{1}{2} - \frac{i}{2}

因为分母相等，所以合并分式:

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

= \frac{- 1 - i}{2}

\frac{- 1 - i}{2}

有理化:

\frac{2 ( - 1 - i )}{2}

\frac{- 1 - i}{2}

乘以共轭根式

\frac{2}{2}

= \frac{( - 1 - i ) 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2 ( - 1 - i )}{2}

= \frac{2 ( - 1 - i )}{2}

将

\frac{2 ( - 1 - i )}{2}

改写成标准复数形式:

- \frac{2}{2} - \frac{2}{2} i

\frac{2 ( - 1 - i )}{2}

使用根式运算法则:

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

= \frac{2 ^{\frac{1}{2}} ( - 1 - i )}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 - i}{2 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 - i}{2 ^{\frac{1}{2}}}

使用根式运算法则:

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

= \frac{- 1 - i}{2}

使用分式法则:

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

\frac{- 1 - i}{2} = - \frac{1}{2} - \frac{i}{2}

= - \frac{1}{2} - \frac{i}{2}

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{1}{2} - \frac{2}{2} i

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2} - \frac{2}{2} i

= - \frac{2}{2} - \frac{2}{2} i

无解

cos (x) = - \frac{1}{2} + \frac{1}{2} i :

无解

cos (x) = - \frac{1}{2} + \frac{1}{2} i

化简

- \frac{1}{2} + \frac{1}{2} i : - \frac{2}{2} + i \frac{2}{2}

- \frac{1}{2} + \frac{1}{2} i

乘

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= - \frac{1}{2} + \frac{i}{2}

因为分母相等，所以合并分式:

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

= \frac{- 1 + i}{2}

\frac{- 1 + i}{2}

有理化:

\frac{2 ( - 1 + i )}{2}

\frac{- 1 + i}{2}

乘以共轭根式

\frac{2}{2}

= \frac{( - 1 + i ) 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2 ( - 1 + i )}{2}

= \frac{2 ( - 1 + i )}{2}

将

\frac{2 ( - 1 + i )}{2}

改写成标准复数形式:

- \frac{2}{2} + \frac{2}{2} i

\frac{2 ( - 1 + i )}{2}

使用根式运算法则:

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

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + i}{2 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + i}{2 ^{\frac{1}{2}}}

使用根式运算法则:

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

= \frac{- 1 + i}{2}

使用分式法则:

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

\frac{- 1 + i}{2} = - \frac{1}{2} + \frac{i}{2}

= - \frac{1}{2} + \frac{i}{2}

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

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= - \frac{1}{2} + \frac{2}{2} i

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2} + \frac{2}{2} i

= - \frac{2}{2} + \frac{2}{2} i

无解

cos (x) = \frac{1}{2} - \frac{1}{2} i :

无解

cos (x) = \frac{1}{2} - \frac{1}{2} i

化简

\frac{1}{2} - \frac{1}{2} i : \frac{2}{2} - i \frac{2}{2}

\frac{1}{2} - \frac{1}{2} i

乘

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

因为分母相等，所以合并分式:

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

= \frac{1 - i}{2}

\frac{1 - i}{2}

有理化:

\frac{2 ( 1 - i )}{2}

\frac{1 - i}{2}

乘以共轭根式

\frac{2}{2}

= \frac{( 1 - i ) 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2 ( 1 - i )}{2}

= \frac{2 ( 1 - i )}{2}

将

\frac{2 ( 1 - i )}{2}

改写成标准复数形式:

\frac{2}{2} - \frac{2}{2} i

\frac{2 ( 1 - i )}{2}

使用根式运算法则:

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

= \frac{2 ^{\frac{1}{2}} ( 1 - i )}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{1 - i}{2 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1 - i}{2 ^{\frac{1}{2}}}

使用根式运算法则:

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

= \frac{1 - i}{2}

使用分式法则:

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

\frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= \frac{1}{2} - \frac{2}{2} i

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

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2} - \frac{2}{2} i

= \frac{2}{2} - \frac{2}{2} i

无解

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

流行的例子

2sin^3(x)-5sin^2(x)+2sin(x)=0

(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

(sin(x)-(sqrt(2)))/2 =0

cos(2x)=5-6cos^2(x)