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受欢迎的三角函数 >

tan(x)+cot(x)=6,sin^6(x)+cos^6(x)

解答

tan (x) + cot (x) = 6, sin^{6} (x) + cos^{6} (x)

解答

x \in R 无解

求解步骤

tan (x) + cot (x) = 6, sin^{6} (x) + cos^{6} (x)

两边减去

6

tan (x) + cot (x) - 6 = 0

使用三角恒等式改写

- 6 + cot (x) + tan (x)

使用基本三角恒等式:

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

= - 6 + cot (x) + \frac{1}{cot ( x )}

- 6 + cot (x) + \frac{1}{cot ( x )} = 0

用替代法求解

- 6 + cot (x) + \frac{1}{cot ( x )} = 0

令

: cot (x) = u

- 6 + u + \frac{1}{u} = 0

- 6 + u + \frac{1}{u} = 0 : u = 3 + 22, u = 3 - 22

- 6 + u + \frac{1}{u} = 0

在两边乘以

u

- 6 + u + \frac{1}{u} = 0

在两边乘以

u

- 6 u + uu + \frac{1}{u} u = 0 \cdot u

化简

- 6 u + uu + \frac{1}{u} u = 0 \cdot u

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

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

约分:

u

= 1

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 6 u + u^{2} + 1 = 0 : u = 3 + 22, u = 3 - 22

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 6)^{2} - 4 \cdot 1 \cdot 1 = 42

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

使用指数法则:

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

若

n

是偶数

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

= 6^{2} - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 6^{2} - 4

6^{2} = 36

= 36 - 4

数字相减:

36 - 4 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

使用指数法则:

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

= 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{4}

使用根式运算法则:

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

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

= 2^{2} 2

整理后得

= 42

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

将解分隔开

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

u = \frac{- ( - 6 ) + 4 2}{2 \cdot 1} : 3 + 22

\frac{- ( - 6 ) + 4 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{6 + 4 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6 + 4 2}{2}

分解

6 + 42 : 2 (3 + 22)

6 + 42

改写为

= 2 \cdot 3 + 2 \cdot 22

因式分解出通项

2

= 2 (3 + 22)

= \frac{2 ( 3 + 2 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 3 + 22

u = \frac{- ( - 6 ) - 4 2}{2 \cdot 1} : 3 - 22

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

使用法则

- (- a) = a

= \frac{6 - 4 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

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

分解

6 - 42 : 2 (3 - 22)

6 - 42

改写为

= 2 \cdot 3 - 2 \cdot 22

因式分解出通项

2

= 2 (3 - 22)

= \frac{2 ( 3 - 2 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 3 - 22

二次方程组的解是:

u = 3 + 22, u = 3 - 22

u = 3 + 22, u = 3 - 22

验证解

找到无定义的点（奇点）:

u = 0

取

- 6 + u + \frac{1}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 3 + 22, u = 3 - 22

u = cot (x)

代回

cot (x) = 3 + 22, cot (x) = 3 - 22

cot (x) = 3 + 22, cot (x) = 3 - 22

cot (x) = 3 + 22, sin^{6} (x) + cos^{6} (x) :

无解

cot (x) = 3 + 22, sin^{6} (x) + cos^{6} (x)

使用反三角函数性质

cot (x) = 3 + 22

cot (x) = 3 + 22

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (3 + 22) + πn

x = arccot (3 + 22) + πn

在

sin^{6} (x) + cos^{6} (x)

范围内的解

无解

cot (x) = 3 - 22, sin^{6} (x) + cos^{6} (x) :

无解

cot (x) = 3 - 22, sin^{6} (x) + cos^{6} (x)

使用反三角函数性质

cot (x) = 3 - 22

cot (x) = 3 - 22

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (3 - 22) + πn

x = arccot (3 - 22) + πn

在

sin^{6} (x) + cos^{6} (x)

范围内的解

无解

合并所有解

x \in R 无解

作图

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