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受欢迎的三角函数 >

1/2 cos^2(2x)+tan(162)=0

解答

\frac{1}{2} cos^{2} (2 x) + tan (16 2^{\circ}) = 0

解答

x = \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = \frac{2.50837 \dots}{2} + 18 0^{\circ} n, x = - \frac{2.50837 \dots}{2} + 18 0^{\circ} n

+1

弧度

x = \frac{0.63322 \dots}{2} + πn, x = π - \frac{0.63322 \dots}{2} + πn, x = \frac{2.50837 \dots}{2} + πn, x = - \frac{2.50837 \dots}{2} + πn

求解步骤

\frac{1}{2} cos^{2} (2 x) + tan (16 2^{\circ}) = 0

tan (16 2^{\circ}) = - \frac{5 - 2 5}{5}

tan (16 2^{\circ})

使用三角恒等式改写:

- tan (1 8^{\circ})

tan (16 2^{\circ})

利用以下特性:

tan (x) = - tan (18 0^{\circ} - x)

tan (x)

利用以下特性:

tan (θ) = - tan (- θ)

tan (x) = - tan (- x)

= - tan (- x)

使用周期

tan

:

tan (18 0^{\circ} + θ) = tan (θ)

- tan (- x) = - tan (18 0^{\circ} - x)

= - tan (18 0^{\circ} - x)

= - tan (36 0^{\circ} - 16 2^{\circ})

化简:

36 0^{\circ} - 16 2^{\circ} = 19 8^{\circ}

36 0^{\circ} - 16 2^{\circ}

将项转换为分式:

36 0^{\circ} = 36 0^{\circ}

= 36 0^{\circ} - 16 2^{\circ}

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{36 0 ^{\circ} 10 - 162 0 ^{\circ}}{10}

36 0^{\circ} 10 - 162 0^{\circ} = 198 0^{\circ}

36 0^{\circ} 10 - 162 0^{\circ}

数字相乘:

2 \cdot 10 = 20

= 360 0^{\circ} - 162 0^{\circ}

同类项相加:

360 0^{\circ} - 162 0^{\circ} = 198 0^{\circ}

= 198 0^{\circ}

= 19 8^{\circ}

= - tan (19 8^{\circ})

tan (19 8^{\circ}) = tan (1 8^{\circ})

tan (19 8^{\circ})

将

19 8^{\circ}

改写为

18 0^{\circ} + 1 8^{\circ}

= tan (18 0^{\circ} + 1 8^{\circ})

使用周期

tan

:

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 1 8^{\circ}) = tan (1 8^{\circ})

= tan (1 8^{\circ})

= - tan (1 8^{\circ})

= - tan (1 8^{\circ})

使用三角恒等式改写:

tan (1 8^{\circ}) = \frac{5 - 2 5}{5}

tan (1 8^{\circ})

使用三角恒等式改写:

\frac{1 - cos ( 3 6 ^{\circ} )}{1 + cos ( 3 6 ^{\circ} )}

tan (1 8^{\circ})

将

tan (1 8^{\circ})

写为

tan (\frac{3 6 ^{\circ}}{2})

= tan (\frac{3 6 ^{\circ}}{2})

使用半角公式:

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

使用三角恒等式改写:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

利用以下特性

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

两边进行平方

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

使用三角恒等式改写:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用三角恒等式改写:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

化简

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

用

\frac{θ}{2}

替代

θ

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{1 + cos ( 2 \cdot \frac{θ}{2} )}

化简

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] quadrant I II tan positive negative

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

= \frac{1 - cos ( 3 6 ^{\circ} )}{1 + cos ( 3 6 ^{\circ} )}

= \frac{1 - cos ( 3 6 ^{\circ} )}{1 + cos ( 3 6 ^{\circ} )}

使用三角恒等式改写:

cos (3 6^{\circ}) = \frac{5 + 1}{4}

cos (3 6^{\circ})

显示:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

显示:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

代入

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

在两侧开平方

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

不能为负

sin (1 8^{\circ})

不能为负

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

以下方程式相加

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}}

化简

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}} : \frac{5 - 2 5}{5}

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}}

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}} = \frac{3 - 5}{5 + 5}

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}}

化简

1 + \frac{5 + 1}{4} : \frac{5 + 5}{4}

1 + \frac{5 + 1}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{5 + 1}{4}

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 4 + 5 + 1}{4}

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

数字相乘:

1 \cdot 4 = 4

= 4 + 5 + 1

数字相加:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

= \frac{1 - \frac{1 + 5}{4}}{\frac{5 + 5}{4}}

化简

1 - \frac{5 + 1}{4} : \frac{3 - 5}{4}

1 - \frac{5 + 1}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{5 + 1}{4}

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 4 - ( 5 + 1 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 - ( 1 + 5 )}{4}

乘开

4 - (5 + 1) : 3 - 5

4 - (5 + 1)

- (5 + 1) : - 5 - 1

- (5 + 1)

打开括号

= - (5) - (1)

使用加减运算法则

+ (- a) = - a

= - 5 - 1

= 4 - 5 - 1

数字相减:

4 - 1 = 3

= 3 - 5

= \frac{3 - 5}{4}

= \frac{\frac{3 - 5}{4}}{\frac{5 + 5}{4}}

分式相除:

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

= \frac{( 3 - 5 ) \cdot 4}{4 ( 5 + 5 )}

约分:

4

= \frac{3 - 5}{5 + 5}

= \frac{3 - 5}{5 + 5}

\frac{3 - 5}{5 + 5} = \frac{5 - 2 5}{5}

\frac{3 - 5}{5 + 5}

乘以共轭根式

\frac{5 - 5}{5 - 5}

= \frac{( 3 - 5 ) ( 5 - 5 )}{( 5 + 5 ) ( 5 - 5 )}

(3 - 5) (5 - 5) = 20 - 85

(3 - 5) (5 - 5)

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 3, b = - 5, c = 5, d = - 5

= 3 \cdot 5 + 3 (- 5) + (- 5) \cdot 5 + (- 5) (- 5)

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= 3 \cdot 5 - 35 - 55 + 55

化简

3 \cdot 5 - 35 - 55 + 55 : 20 - 85

3 \cdot 5 - 35 - 55 + 55

同类项相加:

- 35 - 55 = - 85

= 3 \cdot 5 - 85 + 55

数字相乘:

3 \cdot 5 = 15

= 15 - 85 + 55

使用根式运算法则:

a a = a

55 = 5

= 15 - 85 + 5

数字相加:

15 + 5 = 20

= 20 - 85

= 20 - 85

(5 + 5) (5 - 5) = 20

(5 + 5) (5 - 5)

使用平方差公式:

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

a = 5, b = 5

= 5^{2} - (5)^{2}

化简

5^{2} - (5)^{2} : 20

5^{2} - (5)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

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

= (5^{\frac{1}{2}})^{2}

使用指数法则:

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

= 5^{\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

= 5

= 25 - 5

数字相减:

25 - 5 = 20

= 20

= 20

= \frac{20 - 8 5}{20}

分解

20 - 85 : 4 (5 - 25)

20 - 85

改写为

= 4 \cdot 5 - 4 \cdot 25

因式分解出通项

4

= 4 (5 - 25)

= \frac{4 ( 5 - 2 5 )}{20}

约分:

4

= \frac{5 - 2 5}{5}

= \frac{5 - 2 5}{5}

= \frac{5 - 2 5}{5}

= - \frac{5 - 2 5}{5}

\frac{1}{2} cos^{2} (2 x) + - \frac{5 - 2 5}{5} = 0

使用三角恒等式改写

- \frac{5 - 2 5}{5} + cos^{2} (2 x) \frac{1}{2}

使用基本三角恒等式:

cos (x) = \frac{1}{sec ( x )}

= - \frac{5 - 2 5}{5} + (\frac{1}{sec ( 2 x )})^{2} \frac{1}{2}

化简

- \frac{5 - 2 5}{5} + (\frac{1}{sec ( 2 x )})^{2} \frac{1}{2} : - \frac{5 - 2}{5} + \frac{1}{2 sec ^{2} ( 2 x )}

- \frac{5 - 2 5}{5} + (\frac{1}{sec ( 2 x )})^{2} \frac{1}{2}

\frac{5 - 2 5}{5} = \frac{5 - 2}{5}

\frac{5 - 2 5}{5}

\frac{5 - 2 5}{5} = \frac{5 - 2}{5}

\frac{5 - 2 5}{5}

分解

5 - 25 : 5 (5 - 2)

5 - 25

5 = 55

= 55 - 25

因式分解出通项

5

= 5 (5 - 2)

= \frac{5 ( 5 - 2 )}{5}

消掉

\frac{5 ( 5 - 2 )}{5} : \frac{5 - 2}{5}

\frac{5 ( 5 - 2 )}{5}

使用根式运算法则:

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

5 = 5^{\frac{1}{2}}

= \frac{5 ^{\frac{1}{2}} ( 5 - 2 )}{5}

使用指数法则:

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

\frac{5 ^{\frac{1}{2}}}{5 ^{1}} = \frac{1}{5 ^{1 - \frac{1}{2}}}

= \frac{5 - 2}{5 ^{1 - \frac{1}{2}}}

数字相减:

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

= \frac{5 - 2}{5 ^{\frac{1}{2}}}

使用根式运算法则:

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

5^{\frac{1}{2}} = 5

= \frac{5 - 2}{5}

= \frac{5 - 2}{5}

= \frac{5 - 2}{5}

(\frac{1}{sec ( 2 x )})^{2} \frac{1}{2} = \frac{1}{2 sec ^{2} ( 2 x )}

(\frac{1}{sec ( 2 x )})^{2} \frac{1}{2}

(\frac{1}{sec ( 2 x )})^{2} = \frac{1}{sec ^{2} ( 2 x )}

(\frac{1}{sec ( 2 x )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sec ^{2} ( 2 x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( 2 x )}

= \frac{1}{2} \cdot \frac{1}{sec ^{2} ( 2 x )}

分式相乘:

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

= \frac{1 \cdot 1}{sec ^{2} ( 2 x ) \cdot 2}

数字相乘:

1 \cdot 1 = 1

= \frac{1}{2 sec ^{2} ( 2 x )}

= - \frac{5 - 2}{5} + \frac{1}{2 sec ^{2} ( 2 x )}

= - \frac{5 - 2}{5} + \frac{1}{2 sec ^{2} ( 2 x )}

\frac{1}{2 sec ^{2} ( 2 x )} - \frac{- 2 + 5}{5} = 0

用替代法求解

\frac{1}{2 sec ^{2} ( 2 x )} - \frac{- 2 + 5}{5} = 0

令

: sec (2 x) = u

\frac{1}{2 u ^{2}} - \frac{- 2 + 5}{5} = 0

\frac{1}{2 u ^{2}} - \frac{- 2 + 5}{5} = 0 : u = \frac{( 2 + 5 ) 5 - 2 5}{2}, u = - \frac{( 2 + 5 ) 5 - 2 5}{2}

\frac{1}{2 u ^{2}} - \frac{- 2 + 5}{5} = 0

在两边乘以

2 u^{2}

\frac{1}{2 u ^{2}} - \frac{- 2 + 5}{5} = 0

在两边乘以

2 u^{2}

\frac{1}{2 u ^{2}} \cdot 2 u^{2} - \frac{- 2 + 5}{5} \cdot 2 u^{2} = 0 \cdot 2 u^{2}

化简

\frac{1}{2 u ^{2}} \cdot 2 u^{2} - \frac{- 2 + 5}{5} \cdot 2 u^{2} = 0 \cdot 2 u^{2}

化简

\frac{1}{2 u ^{2}} \cdot 2 u^{2} : 1

\frac{1}{2 u ^{2}} \cdot 2 u^{2}

分式相乘:

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

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

约分:

2

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

约分:

u^{2}

= 1

化简

- \frac{- 2 + 5}{5} \cdot 2 u^{2} : - 2 \frac{- 2 5 + 5}{5} u^{2}

- \frac{- 2 + 5}{5} \cdot 2 u^{2}

\frac{- 2 + 5}{5} = \frac{- 2 5 + 5}{5}

\frac{- 2 + 5}{5}

\frac{- 2 + 5}{5} = \frac{- 2 5 + 5}{5}

\frac{- 2 + 5}{5}

乘以共轭根式

\frac{5}{5}

= \frac{( - 2 + 5 ) 5}{5 5}

(- 2 + 5) 5 = - 25 + 5

(- 2 + 5) 5

= 5 (- 2 + 5)

使用分配律:

a (b + c) = ab + a c

a = 5, b = - 2, c = 5

= 5 (- 2) + 55

使用加减运算法则

+ (- a) = - a

= - 25 + 55

使用根式运算法则:

a a = a

55 = 5

= - 25 + 5

55 = 5

55

使用根式运算法则:

a a = a

55 = 5

= 5

= \frac{- 2 5 + 5}{5}

= \frac{- 2 5 + 5}{5}

= - 2 \frac{5 - 2 5}{5} u^{2}

化简

0 \cdot 2 u^{2} : 0

0 \cdot 2 u^{2}

使用法则

0 \cdot a = 0

= 0

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0

解

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0 : u = \frac{( 2 + 5 ) 5 - 2 5}{2}, u = - \frac{( 2 + 5 ) 5 - 2 5}{2}

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0

将

1

到右边

1 - 2 \frac{- 2 5 + 5}{5} u^{2} = 0

两边减去

1

1 - 2 \frac{- 2 5 + 5}{5} u^{2} - 1 = 0 - 1

化简

- 2 \frac{- 2 5 + 5}{5} u^{2} = - 1

- 2 \frac{- 2 5 + 5}{5} u^{2} = - 1

两边除以

- 2 \frac{- 2 5 + 5}{5}

- 2 \frac{- 2 5 + 5}{5} u^{2} = - 1

两边除以

- 2 \frac{- 2 5 + 5}{5}

\frac{- 2 \frac{- 2 5 + 5}{5} u ^{2}}{- 2 \frac{- 2 5 + 5}{5}} = \frac{- 1}{- 2 \frac{- 2 5 + 5}{5}}

化简

\frac{- 2 \frac{- 2 5 + 5}{5} u ^{2}}{- 2 \frac{- 2 5 + 5}{5}} = \frac{- 1}{- 2 \frac{- 2 5 + 5}{5}}

化简

\frac{- 2 \frac{- 2 5 + 5}{5} u ^{2}}{- 2 \frac{- 2 5 + 5}{5}} : u^{2}

\frac{- 2 \frac{- 2 5 + 5}{5} u ^{2}}{- 2 \frac{- 2 5 + 5}{5}}

使用分式法则:

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

= \frac{2 \frac{- 2 5 + 5}{5} u ^{2}}{2 \frac{- 2 5 + 5}{5}}

数字相除:

\frac{2}{2} = 1

= \frac{\frac{5 - 2 5}{5} u ^{2}}{\frac{- 2 5 + 5}{5}}

约分:

\frac{- 2 5 + 5}{5}

= u^{2}

化简

\frac{- 1}{- 2 \frac{- 2 5 + 5}{5}} : \frac{( 2 + 5 ) 5 - 2 5}{2}

\frac{- 1}{- 2 \frac{- 2 5 + 5}{5}}

使用分式法则:

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

= \frac{1}{2 \frac{- 2 5 + 5}{5}}

\frac{- 2 5 + 5}{5} = \frac{- 2 5 + 5}{5}

\frac{- 2 5 + 5}{5}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{- 2 5 + 5}{5}

= \frac{1}{2 \cdot \frac{5 - 2 5}{5}}

乘

2 \cdot \frac{- 2 5 + 5}{5} : \frac{2 5 - 2 5}{5}

2 \cdot \frac{- 2 5 + 5}{5}

分式相乘:

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

= \frac{- 2 5 + 5 \cdot 2}{5}

= \frac{1}{\frac{2 5 - 2 5}{5}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{5}{- 2 5 + 5 \cdot 2}

\frac{5}{2 5 - 2 5}

有理化:

\frac{( 2 + 5 ) 5 - 2 5}{2}

\frac{5}{2 5 - 2 5}

乘以共轭根式

\frac{- 2 5 + 5}{- 2 5 + 5}

= \frac{5 - 2 5 + 5}{- 2 5 + 5 \cdot 2 - 2 5 + 5}

- 25 + 5 \cdot 2 - 25 + 5 = - 45 + 10

- 25 + 5 \cdot 2 - 25 + 5

使用根式运算法则:

a a = a

5 - 25 5 - 25 = - 25 + 5

= 2 (5 - 25)

使用分配律:

a (b + c) = ab + a c

a = 2, b = - 25, c = 5

= 2 (- 25) + 2 \cdot 5

使用加减运算法则

+ (- a) = - a

= - 2 \cdot 25 + 2 \cdot 5

化简

- 2 \cdot 25 + 2 \cdot 5 : - 45 + 10

- 2 \cdot 25 + 2 \cdot 5

数字相乘:

2 \cdot 2 = 4

= - 45 + 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= - 45 + 10

= - 45 + 10

= \frac{5 - 2 5 + 5}{- 4 5 + 10}

因式分解出通项

- 2 : - 2 (25 - 5)

- 45 + 10

将

10

改写为

2 \cdot 5

将

4

改写为

2 \cdot 2

= - 2 \cdot 25 + 2 \cdot 5

因式分解出通项

- 2

= - 2 (25 - 5)

= - \frac{5 - 2 5 + 5}{2 ( 2 5 - 5 )}

消掉

- \frac{5 - 2 5 + 5}{2 ( 2 5 - 5 )} : \frac{5 - 2 5 + 5}{2 ( 5 - 2 5 )}

- \frac{5 - 2 5 + 5}{2 ( 2 5 - 5 )}

25 - 5 = - (5 - 25)

= - \frac{5 5 - 2 5}{- 2 ( 5 - 2 5 )}

整理后得

= \frac{5 - 2 5 + 5}{2 ( 5 - 2 5 )}

= \frac{5 - 2 5 + 5}{2 ( 5 - 2 5 )}

乘以共轭根式

\frac{5 + 2 5}{5 + 2 5}

= \frac{5 - 2 5 + 5 ( 5 + 2 5 )}{2 ( 5 - 2 5 ) ( 5 + 2 5 )}

5 - 25 + 5 (5 + 25) = 55 - 25 + 5 + 10 - 25 + 5

5 - 25 + 5 (5 + 25)

= 5 (5 + 25) - 25 + 5

使用分配律:

a (b + c) = ab + a c

a = 5 - 25 + 5, b = 5, c = 25

= 5 - 25 + 5 \cdot 5 + 5 - 25 + 5 \cdot 25

= 55 - 25 + 5 + 255 - 25 + 5

255 - 25 + 5 = 10 - 25 + 5

255 - 25 + 5

使用根式运算法则:

a a = a

55 = 5

= 2 \cdot 5 5 - 25

数字相乘:

2 \cdot 5 = 10

= 10 5 - 25

= 55 - 25 + 5 + 10 - 25 + 5

2 (5 - 25) (5 + 25) = 10

2 (5 - 25) (5 + 25)

乘开

(5 - 25) (5 + 25) : 5

(5 - 25) (5 + 25)

使用平方差公式:

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

a = 5, b = 25

= 5^{2} - (25)^{2}

化简

5^{2} - (25)^{2} : 5

5^{2} - (25)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(25)^{2} = 20

(25)^{2}

使用指数法则:

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

= 2^{2} (5)^{2}

(5)^{2} : 5

使用根式运算法则:

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

= (5^{\frac{1}{2}})^{2}

使用指数法则:

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

= 5^{\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

= 5

= 2^{2} \cdot 5

2^{2} = 4

= 4 \cdot 5

数字相乘:

4 \cdot 5 = 20

= 20

= 25 - 20

数字相减:

25 - 20 = 5

= 5

= 5

= 2 \cdot 5

乘开

2 \cdot 5 : 10

2 \cdot 5

打开括号

= 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= 10

= 10

= \frac{5 5 - 2 5 + 5 + 10 - 2 5 + 5}{10}

分解

55 - 25 + 5 + 10 - 25 + 5 : 5 5 - 25 (5 + 2)

55 - 25 + 5 + 10 - 25 + 5

改写为

= 5 5 - 25 5 + 2 \cdot 5 5 - 25

因式分解出通项

5 5 - 25

= 5 5 - 25 (5 + 2)

= \frac{5 5 - 2 5 ( 5 + 2 )}{10}

约分:

5

= \frac{( 2 + 5 ) 5 - 2 5}{2}

= \frac{( 2 + 5 ) 5 - 2 5}{2}

u^{2} = \frac{( 2 + 5 ) 5 - 2 5}{2}

u^{2} = \frac{( 2 + 5 ) 5 - 2 5}{2}

u^{2} = \frac{( 2 + 5 ) 5 - 2 5}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{( 2 + 5 ) 5 - 2 5}{2}, u = - \frac{( 2 + 5 ) 5 - 2 5}{2}

u = \frac{( 2 + 5 ) 5 - 2 5}{2}, u = - \frac{( 2 + 5 ) 5 - 2 5}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1}{2 u ^{2}} - \frac{- 2 + 5}{5}

的分母，令其等于零

解

2 u^{2} = 0 : u = 0

2 u^{2} = 0

两边除以

2

2 u^{2} = 0

两边除以

2

2 u^{2} = 0

两边除以

2

\frac{2 u ^{2}}{2} = \frac{0}{2}

化简

u^{2} = 0

u^{2} = 0

使用法则

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

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = \frac{( 2 + 5 ) 5 - 2 5}{2}, u = - \frac{( 2 + 5 ) 5 - 2 5}{2}

u = sec (2 x)

代回

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}, sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}, sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2} : x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}

使用反三角函数性质

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}

的通解

sec (x) = a \Rightarrow x = arcsec (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arcsec (a) + 36 0^{\circ} n

2 x = arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n, 2 x = 36 0^{\circ} - arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

2 x = arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n, 2 x = 36 0^{\circ} - arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

解

2 x = arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n : x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

2 x = arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

2 x = arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

\frac{2 x}{2} = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

解

2 x = 36 0^{\circ} - arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n : x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

2 x = 36 0^{\circ} - arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

2 x = 36 0^{\circ} - arcsec \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

\frac{2 x}{2} = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2} : x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

使用反三角函数性质

sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

的通解

sec (x) = - a \Rightarrow x = arcsec (- a) + 36 0^{\circ} n, x = - arcsec (- a) + 36 0^{\circ} n

2 x = arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n, 2 x = - arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

2 x = arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n, 2 x = - arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

解

2 x = arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n : x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

2 x = arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

2 x = arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

\frac{2 x}{2} = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

解

2 x = - arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n : x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

2 x = - arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

2 x = - arcsec - \frac{( 2 + 5 ) 5 - 2 5}{2} + 36 0^{\circ} n

两边除以

2

\frac{2 x}{2} = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

合并所有解

x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

以小数形式表示解

x = \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = \frac{2.50837 \dots}{2} + 18 0^{\circ} n, x = - \frac{2.50837 \dots}{2} + 18 0^{\circ} n

作图

流行的例子

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