zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

2tan(60-x)=tan(x)

解答

2 tan (60 - x) = tan (x)

解答

x = - 1.46681 \dots + πn, x = 0.20575 \dots + πn

+1

度数

x = - 84.04237 \dots^{\circ} + 18 0^{\circ} n, x = 11.78914 \dots^{\circ} + 18 0^{\circ} n

求解步骤

2 tan (60 - x) = tan (x)

两边减去

tan (x)

2 tan (60 - x) - tan (x) = 0

使用三角恒等式改写

- tan (x) + 2 tan (60 - x)

使用角差恒等式:

tan (s - t) = \frac{tan ( s ) - tan ( t )}{1 + tan ( s ) tan ( t )}

= - tan (x) + 2 \cdot \frac{tan ( 60 ) - tan ( x )}{1 + tan ( 60 ) tan ( x )}

化简

- tan (x) + 2 \cdot \frac{tan ( 60 ) - tan ( x )}{1 + tan ( 60 ) tan ( x )} : \frac{2 tan ( 60 ) - 3 tan ( x ) - tan ( 60 ) tan ^{2} ( x )}{1 + tan ( 60 ) tan ( x )}

- tan (x) + 2 \cdot \frac{tan ( 60 ) - tan ( x )}{1 + tan ( 60 ) tan ( x )}

乘

2 \cdot \frac{tan ( 60 ) - tan ( x )}{1 + tan ( 60 ) tan ( x )} : \frac{2 ( - tan ( x ) + tan ( 60 ) )}{1 + tan ( 60 ) tan ( x )}

2 \cdot \frac{tan ( 60 ) - tan ( x )}{1 + tan ( 60 ) tan ( x )}

分式相乘:

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

= \frac{( tan ( 60 ) - tan ( x ) ) \cdot 2}{1 + tan ( 60 ) tan ( x )}

= - tan (x) + \frac{2 ( - tan ( x ) + tan ( 60 ) )}{tan ( 60 ) tan ( x ) + 1}

将项转换为分式:

tan (x) = \frac{tan ( x ) ( 1 + tan ( 60 ) tan ( x ) )}{1 + tan ( 60 ) tan ( x )}

= \frac{( tan ( 60 ) - tan ( x ) ) \cdot 2}{1 + tan ( 60 ) tan ( x )} - \frac{tan ( x ) ( 1 + tan ( 60 ) tan ( x ) )}{1 + tan ( 60 ) tan ( 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 ( 60 ) - tan ( x ) ) \cdot 2 - tan ( x ) ( 1 + tan ( 60 ) tan ( x ) )}{1 + tan ( 60 ) tan ( x )}

乘开

(tan (60) - tan (x)) \cdot 2 - tan (x) (1 + tan (60) tan (x)) : 2 tan (60) - 3 tan (x) - tan (60) tan^{2} (x)

(tan (60) - tan (x)) \cdot 2 - tan (x) (1 + tan (60) tan (x))

= 2 (tan (60) - tan (x)) - tan (x) (1 + tan (60) tan (x))

乘开

2 (tan (60) - tan (x)) : 2 tan (60) - 2 tan (x)

2 (tan (60) - tan (x))

使用分配律:

a (b - c) = ab - a c

a = 2, b = tan (60), c = tan (x)

= 2 tan (60) - 2 tan (x)

= 2 tan (60) - 2 tan (x) - tan (x) (1 + tan (60) tan (x))

乘开

- tan (x) (1 + tan (60) tan (x)) : - tan (x) - tan (60) tan^{2} (x)

- tan (x) (1 + tan (60) tan (x))

使用分配律:

a (b + c) = ab + a c

a = - tan (x), b = 1, c = tan (60) tan (x)

= - tan (x) \cdot 1 + (- tan (x)) tan (60) tan (x)

使用加减运算法则

+ (- a) = - a

= - 1 \cdot tan (x) - tan (60) tan (x) tan (x)

化简

- 1 \cdot tan (x) - tan (60) tan (x) tan (x) : - tan (x) - tan (60) tan^{2} (x)

- 1 \cdot tan (x) - tan (60) tan (x) tan (x)

1 \cdot tan (x) = tan (x)

1 \cdot tan (x)

乘以:

1 \cdot tan (x) = tan (x)

= tan (x)

tan (60) tan (x) tan (x) = tan (60) tan^{2} (x)

tan (60) tan (x) tan (x)

使用指数法则:

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

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

= tan (60) tan^{1 + 1} (x)

数字相加:

1 + 1 = 2

= tan (60) tan^{2} (x)

= - tan (x) - tan (60) tan^{2} (x)

= - tan (x) - tan (60) tan^{2} (x)

= 2 tan (60) - 2 tan (x) - tan (x) - tan (60) tan^{2} (x)

同类项相加:

- 2 tan (x) - tan (x) = - 3 tan (x)

= 2 tan (60) - 3 tan (x) - tan (60) tan^{2} (x)

= \frac{2 tan ( 60 ) - 3 tan ( x ) - tan ( 60 ) tan ^{2} ( x )}{1 + tan ( 60 ) tan ( x )}

= \frac{2 tan ( 60 ) - 3 tan ( x ) - tan ( 60 ) tan ^{2} ( x )}{1 + tan ( 60 ) tan ( x )}

\frac{2 tan ( 60 ) - 3 tan ( x ) - tan ( 60 ) tan ^{2} ( x )}{1 + tan ( 60 ) tan ( x )} = 0

用替代法求解

\frac{2 tan ( 60 ) - 3 tan ( x ) - tan ( 60 ) tan ^{2} ( x )}{1 + tan ( 60 ) tan ( x )} = 0

令

: tan (x) = u

\frac{2 tan ( 60 ) - 3 u - tan ( 60 ) u ^{2}}{1 + tan ( 60 ) u} = 0

\frac{2 tan ( 60 ) - 3 u - tan ( 60 ) u ^{2}}{1 + tan ( 60 ) u} = 0 : u = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, u = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

\frac{2 tan ( 60 ) - 3 u - tan ( 60 ) u ^{2}}{1 + tan ( 60 ) u} = 0

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

2 tan (60) - 3 u - tan (60) u^{2} = 0

解

2 tan (60) - 3 u - tan (60) u^{2} = 0 : u = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, u = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

2 tan (60) - 3 u - tan (60) u^{2} = 0

改写成标准形式

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

- tan (60) u^{2} - 3 u + 2 tan (60) = 0

使用求根公式求解

- tan (60) u^{2} - 3 u + 2 tan (60) = 0

二次方程求根公式:

若

a = - tan (60), b = - 3, c = 2 tan (60)

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - tan ( 60 ) ) \cdot 2 tan ( 60 )}{2 ( - tan ( 60 ) )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - tan ( 60 ) ) \cdot 2 tan ( 60 )}{2 ( - tan ( 60 ) )}

(- 3)^{2} - 4 (- tan (60)) \cdot 2 tan (60) = 9 + 8 tan^{2} (60)

(- 3)^{2} - 4 (- tan (60)) \cdot 2 tan (60)

使用法则

- (- a) = a

= (- 3)^{2} + 4 tan (60) \cdot 2 tan (60)

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

(- 3)^{2}

使用指数法则:

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

若

n

是偶数

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

= 3^{2}

4 tan (60) \cdot 2 tan (60) = 8 tan^{2} (60)

4 tan (60) \cdot 2 tan (60)

数字相乘:

4 \cdot 2 = 8

= 8 tan (60) tan (60)

使用指数法则:

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

tan (60) tan (60) = tan^{1 + 1} (60)

= 8 tan^{1 + 1} (60)

数字相加:

1 + 1 = 2

= 8 tan^{2} (60)

= 3^{2} + 8 tan^{2} (60)

3^{2} = 9

= 9 + 8 tan^{2} (60)

u_{1, 2} = \frac{- ( - 3 ) \pm 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )}, u_{2} = \frac{- ( - 3 ) - 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )}

u = \frac{- ( - 3 ) + 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )} : - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}

\frac{- ( - 3 ) + 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )}

去除括号:

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

= \frac{3 + 9 + 8 tan ^{2} ( 60 )}{- 2 tan ( 60 )}

使用分式法则:

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

= - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}

u = \frac{- ( - 3 ) - 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )} : \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

\frac{- ( - 3 ) - 9 + 8 tan ^{2} ( 60 )}{2 ( - tan ( 60 ) )}

去除括号:

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

= \frac{3 - 9 + 8 tan ^{2} ( 60 )}{- 2 tan ( 60 )}

使用分式法则:

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

3 - 9 + 8 tan^{2} (60) = - (8 tan^{2} (60) + 9 - 3)

= \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

二次方程组的解是:

u = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, u = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

u = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, u = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

验证解

找到无定义的点（奇点）:

u = - \frac{1}{tan ( 60 )}

取

\frac{2 tan ( 60 ) - 3 u - tan ( 60 ) u ^{2}}{1 + tan ( 60 ) u}

的分母，令其等于零

解

1 + tan (60) u = 0 : u = - \frac{1}{tan ( 60 )}

1 + tan (60) u = 0

将

1

到右边

1 + tan (60) u = 0

两边减去

1

1 + tan (60) u - 1 = 0 - 1

化简

tan (60) u = - 1

tan (60) u = - 1

两边除以

tan (60)

tan (60) u = - 1

两边除以

tan (60)

\frac{tan ( 60 ) u}{tan ( 60 )} = \frac{- 1}{tan ( 60 )}

化简

u = - \frac{1}{tan ( 60 )}

u = - \frac{1}{tan ( 60 )}

以下点无定义

u = - \frac{1}{tan ( 60 )}

将不在定义域的点与解相综合:

u = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, u = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

u = tan (x)

代回

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}, tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )} : x = arctan (- \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}) + πn

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}

使用反三角函数性质

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}

tan (x) = - \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}) + πn

x = arctan (- \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}) + πn

tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )} : x = arctan (\frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}) + πn

tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

使用反三角函数性质

tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

tan (x) = \frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}) + πn

x = arctan (\frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}) + πn

合并所有解

x = arctan (- \frac{3 + 9 + 8 tan ^{2} ( 60 )}{2 tan ( 60 )}) + πn, x = arctan (\frac{8 tan ^{2} ( 60 ) + 9 - 3}{2 tan ( 60 )}) + πn

以小数形式表示解

x = - 1.46681 \dots + πn, x = 0.20575 \dots + πn

作图

流行的例子

cos (x) = - 0.71

tan (α) = \frac{8}{10}

sin(2x)+sqrt(2)*cos(x)=0

sin (2 x) + 2 \cdot cos (x) = 0

(6.7)/(sin(33))=(5.4)/(sin(A))

\frac{6.7}{sin ( 3 3 ^{\circ} )} = \frac{5.4}{sin ( A )}

2 - 3 sin (θ) = 0