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受欢迎的三角函数 >

tan(x)=(5+cos(x))/(6sin(x)cos(x))

解答

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

解答

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

两边减去

\frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )} = 0

化简

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )} : \frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )}

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

将项转换为分式:

tan (x) = \frac{tan ( x ) 6 sin ( x ) cos ( x )}{6 sin ( x ) cos ( x )}

= \frac{tan ( x ) \cdot 6 sin ( x ) cos ( x )}{6 sin ( x ) cos ( x )} - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

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{tan ( x ) \cdot 6 sin ( x ) cos ( x ) - ( 5 + cos ( x ) )}{6 sin ( x ) cos ( x )}

乘开

tan (x) \cdot 6 sin (x) cos (x) - (5 + cos (x)) : tan (x) \cdot 6 sin (x) cos (x) - 5 - cos (x)

tan (x) \cdot 6 sin (x) cos (x) - (5 + cos (x))

= 6 tan (x) sin (x) cos (x) - (5 + cos (x))

- (5 + cos (x)) : - 5 - cos (x)

- (5 + cos (x))

打开括号

= - (5) - (cos (x))

使用加减运算法则

+ (- a) = - a

= - 5 - cos (x)

= tan (x) \cdot 6 sin (x) cos (x) - 5 - cos (x)

= \frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )}

\frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

6 tan (x) sin (x) cos (x) - 5 - cos (x) = 0

用 sin, cos 表示

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x) = 0

化简

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x) : 6 sin^{2} (x) - 5 - cos (x)

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x)

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) = 6 sin^{2} (x)

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x)

分式相乘:

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

= \frac{sin ( x ) \cdot 6 sin ( x ) cos ( x )}{cos ( x )}

约分:

cos (x)

= sin (x) \cdot 6 sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= 6 sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 6 sin^{2} (x)

= 6 sin^{2} (x) - 5 - cos (x)

6 sin^{2} (x) - 5 - cos (x) = 0

两边加上

cos (x)

6 sin^{2} (x) - 5 = cos (x)

两边进行平方

(6 sin^{2} (x) - 5)^{2} = cos^{2} (x)

两边减去

cos^{2} (x)

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x) = 0

分解

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x) : (6 sin^{2} (x) - 5 + cos (x)) (6 sin^{2} (x) - 5 - cos (x))

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x)

使用平方差公式:

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

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x) = ((6 sin^{2} (x) - 5) + cos (x)) ((6 sin^{2} (x) - 5) - cos (x))

= ((6 sin^{2} (x) - 5) + cos (x)) ((6 sin^{2} (x) - 5) - cos (x))

整理后得

= (6 sin^{2} (x) + cos (x) - 5) (6 sin^{2} (x) - cos (x) - 5)

(6 sin^{2} (x) - 5 + cos (x)) (6 sin^{2} (x) - 5 - cos (x)) = 0

分别求解每个部分

6 sin^{2} (x) - 5 + cos (x) = 0 or 6 sin^{2} (x) - 5 - cos (x) = 0

6 sin^{2} (x) - 5 + cos (x) = 0 : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

6 sin^{2} (x) - 5 + cos (x) = 0

使用三角恒等式改写

- 5 + cos (x) + 6 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 5 + cos (x) + 6 (1 - cos^{2} (x))

化简

- 5 + cos (x) + 6 (1 - cos^{2} (x)) : cos (x) - 6 cos^{2} (x) + 1

- 5 + cos (x) + 6 (1 - cos^{2} (x))

乘开

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数字相乘:

6 \cdot 1 = 6

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

= - 5 + cos (x) + 6 - 6 cos^{2} (x)

化简

- 5 + cos (x) + 6 - 6 cos^{2} (x) : cos (x) - 6 cos^{2} (x) + 1

- 5 + cos (x) + 6 - 6 cos^{2} (x)

对同类项分组

= cos (x) - 6 cos^{2} (x) - 5 + 6

数字相加/相减:

- 5 + 6 = 1

= cos (x) - 6 cos^{2} (x) + 1

= cos (x) - 6 cos^{2} (x) + 1

= cos (x) - 6 cos^{2} (x) + 1

1 + cos (x) - 6 cos^{2} (x) = 0

用替代法求解

1 + cos (x) - 6 cos^{2} (x) = 0

令

: cos (x) = u

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

1 + u - 6 u^{2} = 0 : u = - \frac{1}{3}, u = \frac{1}{2}

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

改写成标准形式

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

- 6 u^{2} + u + 1 = 0

使用求根公式求解

- 6 u^{2} + u + 1 = 0

二次方程求根公式:

若

a = - 6, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

1^{2} - 4 (- 6) \cdot 1 = 5

1^{2} - 4 (- 6) \cdot 1

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 ( - 6 )}, u_{2} = \frac{- 1 - 5}{2 ( - 6 )}

u = \frac{- 1 + 5}{2 ( - 6 )} : - \frac{1}{3}

\frac{- 1 + 5}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 6}

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4}{- 12}

使用分式法则:

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

= - \frac{4}{12}

约分:

4

= - \frac{1}{3}

u = \frac{- 1 - 5}{2 ( - 6 )} : \frac{1}{2}

\frac{- 1 - 5}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 6}

数字相减:

- 1 - 5 = - 6

= \frac{- 6}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

u = - \frac{1}{3}, u = \frac{1}{2}

u = cos (x)

代回

cos (x) = - \frac{1}{3}, cos (x) = \frac{1}{2}

cos (x) = - \frac{1}{3}, cos (x) = \frac{1}{2}

cos (x) = - \frac{1}{3} : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

cos (x) = - \frac{1}{3}

使用反三角函数性质

cos (x) = - \frac{1}{3}

cos (x) = - \frac{1}{3}

的通解

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

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

合并所有解

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

6 sin^{2} (x) - 5 - cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

6 sin^{2} (x) - 5 - cos (x) = 0

使用三角恒等式改写

- 5 - cos (x) + 6 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 5 - cos (x) + 6 (1 - cos^{2} (x))

化简

- 5 - cos (x) + 6 (1 - cos^{2} (x)) : - 6 cos^{2} (x) - cos (x) + 1

- 5 - cos (x) + 6 (1 - cos^{2} (x))

乘开

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数字相乘:

6 \cdot 1 = 6

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

= - 5 - cos (x) + 6 - 6 cos^{2} (x)

化简

- 5 - cos (x) + 6 - 6 cos^{2} (x) : - 6 cos^{2} (x) - cos (x) + 1

- 5 - cos (x) + 6 - 6 cos^{2} (x)

对同类项分组

= - cos (x) - 6 cos^{2} (x) - 5 + 6

数字相加/相减:

- 5 + 6 = 1

= - 6 cos^{2} (x) - cos (x) + 1

= - 6 cos^{2} (x) - cos (x) + 1

= - 6 cos^{2} (x) - cos (x) + 1

1 - cos (x) - 6 cos^{2} (x) = 0

用替代法求解

1 - cos (x) - 6 cos^{2} (x) = 0

令

: cos (x) = u

1 - u - 6 u^{2} = 0

1 - u - 6 u^{2} = 0 : u = - \frac{1}{2}, u = \frac{1}{3}

1 - u - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} - u + 1 = 0

使用求根公式求解

- 6 u^{2} - u + 1 = 0

二次方程求根公式:

若

a = - 6, b = - 1, c = 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

(- 1)^{2} - 4 (- 6) \cdot 1 = 5

(- 1)^{2} - 4 (- 6) \cdot 1

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \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 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 6 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 6 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 1 ) + 5}{2 ( - 6 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 + 5}{- 2 \cdot 6}

数字相加:

1 + 5 = 6

= \frac{6}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 6 )} : \frac{1}{3}

\frac{- ( - 1 ) - 5}{2 ( - 6 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 - 5}{- 2 \cdot 6}

数字相减:

1 - 5 = - 4

= \frac{- 4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

使用分式法则:

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

= \frac{4}{12}

约分:

4

= \frac{1}{3}

二次方程组的解是:

u = - \frac{1}{2}, u = \frac{1}{3}

u = cos (x)

代回

cos (x) = - \frac{1}{2}, cos (x) = \frac{1}{3}

cos (x) = - \frac{1}{2}, cos (x) = \frac{1}{3}

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

使用反三角函数性质

cos (x) = \frac{1}{3}

cos (x) = \frac{1}{3}

的通解

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

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

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

合并所有解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

合并所有解

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

将解代入原方程进行验证

将它们代入

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{1}{3}) + 2 πn

的解:

假

arccos (- \frac{1}{3}) + 2 πn

代入

n = 1

arccos (- \frac{1}{3}) + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = arccos (- \frac{1}{3}) + 2 π 1

tan (arccos (- \frac{1}{3}) + 2 π 1) = \frac{5 + cos ( arccos ( - \frac{1}{3} ) + 2 π 1 )}{6 sin ( arccos ( - \frac{1}{3} ) + 2 π 1 ) cos ( arccos ( - \frac{1}{3} ) + 2 π 1 )}

整理后得

- 2.82842 \dots = - 2.47487 \dots

\Rightarrow 假

检验

- arccos (- \frac{1}{3}) + 2 πn

的解:

假

- arccos (- \frac{1}{3}) + 2 πn

代入

n = 1

- arccos (- \frac{1}{3}) + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = - arccos (- \frac{1}{3}) + 2 π 1

tan (- arccos (- \frac{1}{3}) + 2 π 1) = \frac{5 + cos ( - arccos ( - \frac{1}{3} ) + 2 π 1 )}{6 sin ( - arccos ( - \frac{1}{3} ) + 2 π 1 ) cos ( - arccos ( - \frac{1}{3} ) + 2 π 1 )}

整理后得

2.82842 \dots = 2.47487 \dots

\Rightarrow 假

检验

\frac{π}{3} + 2 πn

的解:

假

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = \frac{π}{3} + 2 π 1

tan (\frac{π}{3} + 2 π 1) = \frac{5 + cos ( \frac{π}{3} + 2 π 1 )}{6 sin ( \frac{π}{3} + 2 π 1 ) cos ( \frac{π}{3} + 2 π 1 )}

整理后得

1.73205 \dots = 2.11695 \dots

\Rightarrow 假

检验

\frac{5 π}{3} + 2 πn

的解:

假

\frac{5 π}{3} + 2 πn

代入

n = 1

\frac{5 π}{3} + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = \frac{5 π}{3} + 2 π 1

tan (\frac{5 π}{3} + 2 π 1) = \frac{5 + cos ( \frac{5 π}{3} + 2 π 1 )}{6 sin ( \frac{5 π}{3} + 2 π 1 ) cos ( \frac{5 π}{3} + 2 π 1 )}

整理后得

- 1.73205 \dots = - 2.11695 \dots

\Rightarrow 假

检验

\frac{2 π}{3} + 2 πn

的解:

真

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = \frac{2 π}{3} + 2 π 1

tan (\frac{2 π}{3} + 2 π 1) = \frac{5 + cos ( \frac{2 π}{3} + 2 π 1 )}{6 sin ( \frac{2 π}{3} + 2 π 1 ) cos ( \frac{2 π}{3} + 2 π 1 )}

整理后得

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

检验

\frac{4 π}{3} + 2 πn

的解:

真

\frac{4 π}{3} + 2 πn

代入

n = 1

\frac{4 π}{3} + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = \frac{4 π}{3} + 2 π 1

tan (\frac{4 π}{3} + 2 π 1) = \frac{5 + cos ( \frac{4 π}{3} + 2 π 1 )}{6 sin ( \frac{4 π}{3} + 2 π 1 ) cos ( \frac{4 π}{3} + 2 π 1 )}

整理后得

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

检验

arccos (\frac{1}{3}) + 2 πn

的解:

真

arccos (\frac{1}{3}) + 2 πn

代入

n = 1

arccos (\frac{1}{3}) + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = arccos (\frac{1}{3}) + 2 π 1

tan (arccos (\frac{1}{3}) + 2 π 1) = \frac{5 + cos ( arccos ( \frac{1}{3} ) + 2 π 1 )}{6 sin ( arccos ( \frac{1}{3} ) + 2 π 1 ) cos ( arccos ( \frac{1}{3} ) + 2 π 1 )}

整理后得

2.82842 \dots = 2.82842 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{1}{3}) + 2 πn

的解:

真

2 π - arccos (\frac{1}{3}) + 2 πn

代入

n = 1

2 π - arccos (\frac{1}{3}) + 2 π 1

对于

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

代入

x = 2 π - arccos (\frac{1}{3}) + 2 π 1

tan (2 π - arccos (\frac{1}{3}) + 2 π 1) = \frac{5 + cos ( 2 π - arccos ( \frac{1}{3} ) + 2 π 1 )}{6 sin ( 2 π - arccos ( \frac{1}{3} ) + 2 π 1 ) cos ( 2 π - arccos ( \frac{1}{3} ) + 2 π 1 )}

整理后得

- 2.82842 \dots = - 2.82842 \dots

\Rightarrow 真

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

以小数形式表示解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

作图

流行的例子

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