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受欢迎的三角函数 >

cos(x)(18-12sqrt(2))=12+6sqrt(3)

解答

cos (x) (18 - 122) = 12 + 63

解答

x \in R 无解

求解步骤

cos (x) (18 - 122) = 12 + 63

两边除以

18 - 122

cos (x) (18 - 122) = 12 + 63

两边除以

18 - 122

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} = \frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

化简

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} = \frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

化简

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} : cos (x)

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2}

约分:

18 - 122

= cos (x)

化简

\frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2} : (2 + 3) (3 + 22)

\frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

使用法则

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

= \frac{12 + 6 3}{18 - 12 2}

分解

12 + 63 : 6 (2 + 3)

12 + 63

改写为

= 6 \cdot 2 + 63

因式分解出通项

6

= 6 (2 + 3)

= \frac{6 ( 2 + 3 )}{18 - 12 2}

分解

18 - 122 : 6 (3 - 22)

18 - 122

改写为

= 6 \cdot 3 - 6 \cdot 22

因式分解出通项

6

= 6 (3 - 22)

= \frac{6 ( 2 + 3 )}{6 ( 3 - 2 2 )}

数字相除:

\frac{6}{6} = 1

= \frac{2 + 3}{( 3 - 2 2 )}

去除括号:

(a) = a

= \frac{2 + 3}{3 - 2 2}

\frac{2 + 3}{3 - 2 2}

有理化:

(2 + 3) (3 + 22)

\frac{2 + 3}{3 - 2 2}

乘以共轭根式

\frac{3 + 2 2}{3 + 2 2}

= \frac{( 2 + 3 ) ( 3 + 2 2 )}{( 3 - 2 2 ) ( 3 + 2 2 )}

(3 - 22) (3 + 22) = 1

(3 - 22) (3 + 22)

22 = 2^{\frac{3}{2}}

22

使用指数法则:

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

22 = 2 \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2}}

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

化简

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

将项转换为分式:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

= 2^{\frac{3}{2}}

= (3 - 2^{\frac{3}{2}}) (3 + 22)

22 = 2^{\frac{3}{2}}

22

使用指数法则:

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

22 = 2 \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2}}

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

化简

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

将项转换为分式:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

= 2^{\frac{3}{2}}

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

使用平方差公式:

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

a = 3, b = 2^{\frac{3}{2}}

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

化简

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

3^{2} - (2^{\frac{3}{2}})^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(2^{\frac{3}{2}})^{2} = 8

(2^{\frac{3}{2}})^{2}

使用指数法则:

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

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

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

\frac{3}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 3

= 2^{3}

2^{3} = 8

= 8

= 9 - 8

数字相减:

9 - 8 = 1

= 1

= 1

= \frac{( 2 + 3 ) ( 3 + 2 2 )}{1}

使用法则

\frac{a}{1} = a

= (2 + 3) (3 + 22)

= (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

- 1 \leq cos (x) \leq 1

x \in R 无解

作图

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