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受欢迎的三角函数 >

4tan(2x)-4cot(x)=0

解答

4 tan (2 x) - 4 cot (x) = 0

解答

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

+1

度数

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

求解步骤

4 tan (2 x) - 4 cot (x) = 0

使用三角恒等式改写

- 4 cot (x) + 4 tan (2 x)

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

tan (2 x)

使用倍角公式:

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

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

分解

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

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

分解

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

1 - tan^{2} (x)

使用平方差公式:

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

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

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

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

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

= - 4 cot (x) + 4 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

4 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} = \frac{8 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

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

分式相乘:

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

= \frac{2 tan ( x ) \cdot 4}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

数字相乘:

2 \cdot 4 = 8

= \frac{8 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )}

= - 4 cot (x) + \frac{8 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

使用基本三角恒等式:

cot (x) = \frac{1}{tan ( x )}

= \frac{8 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 4 \cdot \frac{1}{tan ( x )}

化简

\frac{8 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 4 \cdot \frac{1}{tan ( x )} : - \frac{12 tan ^{2} ( x ) - 4}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

\frac{8 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 4 \cdot \frac{1}{tan ( x )}

4 \cdot \frac{1}{tan ( x )} = \frac{4}{tan ( x )}

4 \cdot \frac{1}{tan ( x )}

分式相乘:

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

= \frac{1 \cdot 4}{tan ( x )}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{tan ( x )}

= \frac{8 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )} - \frac{4}{tan ( x )}

分解

(1 + tan (x)) (1 - tan (x)) : - (1 + tan (x)) (tan (x) - 1)

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

分解

1 - tan (x) : - (tan (x) - 1)

1 - tan (x)

因式分解出通项

- 1

= - (tan (x) - 1)

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

= \frac{8 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} - \frac{4}{tan ( x )}

- (1 + tan (x)) (tan (x) - 1), tan (x)

的最小公倍数:

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

- (1 + tan (x)) (tan (x) - 1), tan (x)

最小公倍数 (LCM)

计算出由出现在

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

或

tan (x)

中的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

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

对于

\frac{8 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} :

将分母和分子乘以

tan (x)

\frac{8 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} = \frac{8 tan ( x ) tan ( x )}{( - ( 1 + tan ( x ) ) ( tan ( x ) - 1 ) ) tan ( x )} = \frac{8 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

对于

\frac{4}{tan ( x )} :

将分母和分子乘以

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

\frac{4}{tan ( x )} = \frac{4 ( - ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )}{tan ( x ) ( - ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )} = \frac{- 4 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

= \frac{8 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )} - \frac{- 4 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

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{8 tan ^{2} ( x ) - ( - 4 ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

整理后得

= - \frac{8 tan ^{2} ( x ) + 4 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

乘开

8 tan^{2} (x) + 4 (tan (x) + 1) (tan (x) - 1) : 12 tan^{2} (x) - 4

8 tan^{2} (x) + 4 (tan (x) + 1) (tan (x) - 1)

乘开

4 (tan (x) + 1) (tan (x) - 1) : 4 tan^{2} (x) - 4

乘开

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

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

使用平方差公式:

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

a = tan (x), b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= tan^{2} (x) - 1

= 4 (tan^{2} (x) - 1)

乘开

4 (tan^{2} (x) - 1) : 4 tan^{2} (x) - 4

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = tan^{2} (x), c = 1

= 4 tan^{2} (x) - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 4 tan^{2} (x) - 4

= 4 tan^{2} (x) - 4

= 8 tan^{2} (x) + 4 tan^{2} (x) - 4

同类项相加:

8 tan^{2} (x) + 4 tan^{2} (x) = 12 tan^{2} (x)

= 12 tan^{2} (x) - 4

= - \frac{12 tan ^{2} ( x ) - 4}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

= - \frac{12 tan ^{2} ( x ) - 4}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

- \frac{- 4 + 12 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

用替代法求解

- \frac{- 4 + 12 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

令

: tan (x) = u

- \frac{- 4 + 12 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

- \frac{- 4 + 12 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- \frac{- 4 + 12 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

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

- 4 + 12 u^{2} = 0

解

- 4 + 12 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- 4 + 12 u^{2} = 0

将

4

到右边

- 4 + 12 u^{2} = 0

两边加上

4

- 4 + 12 u^{2} + 4 = 0 + 4

化简

12 u^{2} = 4

12 u^{2} = 4

两边除以

12

12 u^{2} = 4

两边除以

12

\frac{12 u ^{2}}{12} = \frac{4}{12}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

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

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

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

验证解

找到无定义的点（奇点）:

u = 1, u = - 1, u = 0

取

- \frac{- 4 + 12 u ^{2}}{( - 1 + u ) ( 1 + u ) u}

的分母，令其等于零

解

(- 1 + u) (1 + u) u = 0 : u = 1, u = - 1, u = 0

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

- 1 + u = 0 or 1 + u = 0 or u = 0

解

- 1 + u = 0 : u = 1

- 1 + u = 0

将

1

到右边

- 1 + u = 0

两边加上

1

- 1 + u + 1 = 0 + 1

化简

u = 1

u = 1

解

1 + u = 0 : u = - 1

1 + u = 0

将

1

到右边

1 + u = 0

两边减去

1

1 + u - 1 = 0 - 1

化简

u = - 1

u = - 1

解为

u = 1, u = - 1, u = 0

以下点无定义

u = 1, u = - 1, u = 0

将不在定义域的点与解相综合:

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

u = tan (x)

代回

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

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

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

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

使用反三角函数性质

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

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

的通解

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

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

使用反三角函数性质

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

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

的通解

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

合并所有解

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

以小数形式表示解

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

作图

流行的例子

sin (x) = \frac{24}{25}

arcsin(3x-pi)= 1/2

arcsin (3 x - π) = \frac{1}{2}

2cos^2(θ)+sin(θ)-1=0

2 cos^{2} (θ) + sin (θ) - 1 = 0

2sin(2x)sin(x)-3cos(x)=0

2 sin (2 x) sin (x) - 3 cos (x) = 0

sin(θ)=(sqrt(13))/7

sin (θ) = \frac{13}{7}