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tan(x)=-24/7 tan(2x)

解答

tan (x) = - \frac{24}{7} tan (2 x)

解答

x = πn, x = - 1.22811 \dots + πn, x = 1.22811 \dots + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = - 70.36598 \dots^{\circ} + 18 0^{\circ} n, x = 70.36598 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) = - \frac{24}{7} tan (2 x)

两边减去

- \frac{24}{7} tan (2 x)

tan (x) + \frac{24}{7} tan (2 x) = 0

化简

tan (x) + \frac{24}{7} tan (2 x) : \frac{7 tan ( x ) + 24 tan ( 2 x )}{7}

tan (x) + \frac{24}{7} tan (2 x)

乘

\frac{24}{7} tan (2 x) : \frac{24 tan ( 2 x )}{7}

\frac{24}{7} tan (2 x)

分式相乘:

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

= \frac{24 tan ( 2 x )}{7}

= tan (x) + \frac{24 tan ( 2 x )}{7}

将项转换为分式:

tan (x) = \frac{tan ( x ) 7}{7}

= \frac{tan ( x ) \cdot 7}{7} + \frac{24 tan ( 2 x )}{7}

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 7 + 24 tan ( 2 x )}{7}

\frac{7 tan ( x ) + 24 tan ( 2 x )}{7} = 0

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

7 tan (x) + 24 tan (2 x) = 0

使用三角恒等式改写

24 tan (2 x) + 7 tan (x)

使用倍角公式:

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

= 24 \cdot \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + 7 tan (x)

化简

24 \cdot \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + 7 tan (x) : \frac{55 tan ( x ) - 7 tan ^{3} ( x )}{1 - tan ^{2} ( x )}

24 \cdot \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + 7 tan (x)

24 \cdot \frac{2 tan ( x )}{1 - tan ^{2} ( x )} = \frac{48 tan ( x )}{1 - tan ^{2} ( x )}

24 \cdot \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

分式相乘:

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

= \frac{2 tan ( x ) \cdot 24}{1 - tan ^{2} ( x )}

数字相乘:

2 \cdot 24 = 48

= \frac{48 tan ( x )}{1 - tan ^{2} ( x )}

= \frac{48 tan ( x )}{- tan ^{2} ( x ) + 1} + 7 tan (x)

将项转换为分式:

7 tan (x) = \frac{7 tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

= \frac{48 tan ( x )}{1 - tan ^{2} ( x )} + \frac{7 tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( 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{48 tan ( x ) + 7 tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

乘开

48 tan (x) + 7 tan (x) (1 - tan^{2} (x)) : 55 tan (x) - 7 tan^{3} (x)

48 tan (x) + 7 tan (x) (1 - tan^{2} (x))

乘开

7 tan (x) (1 - tan^{2} (x)) : 7 tan (x) - 7 tan^{3} (x)

7 tan (x) (1 - tan^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 7 tan (x), b = 1, c = tan^{2} (x)

= 7 tan (x) \cdot 1 - 7 tan (x) tan^{2} (x)

= 7 \cdot 1 \cdot tan (x) - 7 tan^{2} (x) tan (x)

化简

7 \cdot 1 \cdot tan (x) - 7 tan^{2} (x) tan (x) : 7 tan (x) - 7 tan^{3} (x)

7 \cdot 1 \cdot tan (x) - 7 tan^{2} (x) tan (x)

7 \cdot 1 \cdot tan (x) = 7 tan (x)

7 \cdot 1 \cdot tan (x)

数字相乘:

7 \cdot 1 = 7

= 7 tan (x)

7 tan^{2} (x) tan (x) = 7 tan^{3} (x)

7 tan^{2} (x) tan (x)

使用指数法则:

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

tan^{2} (x) tan (x) = tan^{2 + 1} (x)

= 7 tan^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 7 tan^{3} (x)

= 7 tan (x) - 7 tan^{3} (x)

= 7 tan (x) - 7 tan^{3} (x)

= 48 tan (x) + 7 tan (x) - 7 tan^{3} (x)

同类项相加:

48 tan (x) + 7 tan (x) = 55 tan (x)

= 55 tan (x) - 7 tan^{3} (x)

= \frac{55 tan ( x ) - 7 tan ^{3} ( x )}{1 - tan ^{2} ( x )}

= \frac{55 tan ( x ) - 7 tan ^{3} ( x )}{1 - tan ^{2} ( x )}

\frac{55 tan ( x ) - 7 tan ^{3} ( x )}{1 - tan ^{2} ( x )} = 0

用替代法求解

\frac{55 tan ( x ) - 7 tan ^{3} ( x )}{1 - tan ^{2} ( x )} = 0

令

: tan (x) = u

\frac{55 u - 7 u ^{3}}{1 - u ^{2}} = 0

\frac{55 u - 7 u ^{3}}{1 - u ^{2}} = 0 : u = 0, u = - \frac{55}{7}, u = \frac{55}{7}

\frac{55 u - 7 u ^{3}}{1 - u ^{2}} = 0

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

55 u - 7 u^{3} = 0

解

55 u - 7 u^{3} = 0 : u = 0, u = - \frac{55}{7}, u = \frac{55}{7}

55 u - 7 u^{3} = 0

因式分解

55 u - 7 u^{3} : - u (7 u + 55) (7 u - 55)

55 u - 7 u^{3}

因式分解出通项

- u : - u (7 u^{2} - 55)

- 7 u^{3} + 55 u

使用指数法则:

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

u^{3} = u^{2} u

= - 7 u^{2} u + 55 u

因式分解出通项

- u

= - u (7 u^{2} - 55)

= - u (7 u^{2} - 55)

分解

7 u^{2} - 55 : (7 u + 55) (7 u - 55)

7 u^{2} - 55

将

7 u^{2} - 55

改写为

(7 u)^{2} - (55)^{2}

7 u^{2} - 55

使用根式运算法则:

a = (a)^{2}

7 = (7)^{2}

= (7)^{2} u^{2} - 55

使用根式运算法则:

a = (a)^{2}

55 = (55)^{2}

= (7)^{2} u^{2} - (55)^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(7)^{2} u^{2} = (7 u)^{2}

= (7 u)^{2} - (55)^{2}

= (7 u)^{2} - (55)^{2}

使用平方差公式:

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

(7 u)^{2} - (55)^{2} = (7 u + 55) (7 u - 55)

= (7 u + 55) (7 u - 55)

= - u (7 u + 55) (7 u - 55)

- u (7 u + 55) (7 u - 55) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 7 u + 55 = 0 or 7 u - 55 = 0

解

7 u + 55 = 0 : u = - \frac{55}{7}

7 u + 55 = 0

将

55

到右边

7 u + 55 = 0

两边减去

55

7 u + 55 - 55 = 0 - 55

化简

7 u = - 55

7 u = - 55

两边除以

7

7 u = - 55

两边除以

7

\frac{7 u}{7} = \frac{- 55}{7}

化简

\frac{7 u}{7} = \frac{- 55}{7}

化简

\frac{7 u}{7} : u

\frac{7 u}{7}

约分:

7

= u

化简

\frac{- 55}{7} : - \frac{55}{7}

\frac{- 55}{7}

使用分式法则:

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

= - \frac{55}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= - \frac{55}{7}

u = - \frac{55}{7}

u = - \frac{55}{7}

u = - \frac{55}{7}

解

7 u - 55 = 0 : u = \frac{55}{7}

7 u - 55 = 0

将

55

到右边

7 u - 55 = 0

两边加上

55

7 u - 55 + 55 = 0 + 55

化简

7 u = 55

7 u = 55

两边除以

7

7 u = 55

两边除以

7

\frac{7 u}{7} = \frac{55}{7}

化简

\frac{7 u}{7} = \frac{55}{7}

化简

\frac{7 u}{7} : u

\frac{7 u}{7}

约分:

7

= u

化简

\frac{55}{7} : \frac{55}{7}

\frac{55}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{55}{7}

u = \frac{55}{7}

u = \frac{55}{7}

u = \frac{55}{7}

解为

u = 0, u = - \frac{55}{7}, u = \frac{55}{7}

u = 0, u = - \frac{55}{7}, u = \frac{55}{7}

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

\frac{55 u - 7 u ^{3}}{1 - u ^{2}}

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

u = 0, u = - \frac{55}{7}, u = \frac{55}{7}

u = tan (x)

代回

tan (x) = 0, tan (x) = - \frac{55}{7}, tan (x) = \frac{55}{7}

tan (x) = 0, tan (x) = - \frac{55}{7}, tan (x) = \frac{55}{7}

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - \frac{55}{7} : x = arctan (- \frac{55}{7}) + πn

tan (x) = - \frac{55}{7}

使用反三角函数性质

tan (x) = - \frac{55}{7}

tan (x) = - \frac{55}{7}

的通解

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

x = arctan (- \frac{55}{7}) + πn

x = arctan (- \frac{55}{7}) + πn

tan (x) = \frac{55}{7} : x = arctan (\frac{55}{7}) + πn

tan (x) = \frac{55}{7}

使用反三角函数性质

tan (x) = \frac{55}{7}

tan (x) = \frac{55}{7}

的通解

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

x = arctan (\frac{55}{7}) + πn

x = arctan (\frac{55}{7}) + πn

合并所有解

x = πn, x = arctan (- \frac{55}{7}) + πn, x = arctan (\frac{55}{7}) + πn

以小数形式表示解

x = πn, x = - 1.22811 \dots + πn, x = 1.22811 \dots + πn

流行的例子

tan (x) = \frac{13}{9}

cos(A)= 6/(7*21)

cos (A) = \frac{6}{7 \cdot 21}

2 = 4 cos (3 x + 1)

tan (x) = \frac{13}{6}

3 sinh (2 x) = 5