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受欢迎的三角函数 >

8sin(x)=cos(x)-6

解答

8 sin (x) = cos (x) - 6

解答

x = - 2.17788 \dots + 2 πn, x = 2 π - 0.71499 \dots + 2 πn

+1

度数

x = - 124.78383 \dots^{\circ} + 36 0^{\circ} n, x = 319.03386 \dots^{\circ} + 36 0^{\circ} n

求解步骤

8 sin (x) = cos (x) - 6

两边进行平方

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

两边减去

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

64 sin^{2} (x) - cos^{2} (x) + 12 cos (x) - 36 = 0

使用三角恒等式改写

- 36 - cos^{2} (x) + 12 cos (x) + 64 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)

= - 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x))

化简

- 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x)) : 12 cos (x) - 65 cos^{2} (x) + 28

- 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x))

乘开

64 (1 - cos^{2} (x)) : 64 - 64 cos^{2} (x)

64 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (x)

= 64 \cdot 1 - 64 cos^{2} (x)

数字相乘:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (x)

= - 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x)

化简

- 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x) : 12 cos (x) - 65 cos^{2} (x) + 28

- 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x)

对同类项分组

= - cos^{2} (x) + 12 cos (x) - 64 cos^{2} (x) - 36 + 64

同类项相加:

- cos^{2} (x) - 64 cos^{2} (x) = - 65 cos^{2} (x)

= - 65 cos^{2} (x) + 12 cos (x) - 36 + 64

数字相加/相减:

- 36 + 64 = 28

= 12 cos (x) - 65 cos^{2} (x) + 28

= 12 cos (x) - 65 cos^{2} (x) + 28

= 12 cos (x) - 65 cos^{2} (x) + 28

28 + 12 cos (x) - 65 cos^{2} (x) = 0

用替代法求解

28 + 12 cos (x) - 65 cos^{2} (x) = 0

令

: cos (x) = u

28 + 12 u - 65 u^{2} = 0

28 + 12 u - 65 u^{2} = 0 : u = - \frac{2 ( 4 29 - 3 )}{65}, u = \frac{2 ( 3 + 4 29 )}{65}

28 + 12 u - 65 u^{2} = 0

改写成标准形式

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

- 65 u^{2} + 12 u + 28 = 0

使用求根公式求解

- 65 u^{2} + 12 u + 28 = 0

二次方程求根公式:

若

a = - 65, b = 12, c = 28

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 65 ) \cdot 28}{2 ( - 65 )}

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 65 ) \cdot 28}{2 ( - 65 )}

1 2^{2} - 4 (- 65) \cdot 28 = 1629

1 2^{2} - 4 (- 65) \cdot 28

使用法则

- (- a) = a

= 1 2^{2} + 4 \cdot 65 \cdot 28

数字相乘:

4 \cdot 65 \cdot 28 = 7280

= 1 2^{2} + 7280

1 2^{2} = 144

= 144 + 7280

数字相加:

144 + 7280 = 7424

= 7424

7424

质因数分解:

2^{8} \cdot 29

7424

7424

除以

27424 = 3712 \cdot 2

= 2 \cdot 3712

3712

除以

23712 = 1856 \cdot 2

= 2 \cdot 2 \cdot 1856

1856

除以

21856 = 928 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 928

928

除以

2928 = 464 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 464

464

除以

2464 = 232 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 232

232

除以

2232 = 116 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 116

116

除以

2116 = 58 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 58

58

除以

258 = 29 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 29

2, 29

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 29

= 2^{8} \cdot 29

= 2^{8} \cdot 29

使用根式运算法则:

n ab = n a n b

= 29 2^{8}

使用根式运算法则:

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

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 29

整理后得

= 1629

u_{1, 2} = \frac{- 12 \pm 16 29}{2 ( - 65 )}

将解分隔开

u_{1} = \frac{- 12 + 16 29}{2 ( - 65 )}, u_{2} = \frac{- 12 - 16 29}{2 ( - 65 )}

u = \frac{- 12 + 16 29}{2 ( - 65 )} : - \frac{2 ( 4 29 - 3 )}{65}

\frac{- 12 + 16 29}{2 ( - 65 )}

去除括号:

(- a) = - a

= \frac{- 12 + 16 29}{- 2 \cdot 65}

数字相乘:

2 \cdot 65 = 130

= \frac{- 12 + 16 29}{- 130}

使用分式法则:

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

= - \frac{- 12 + 16 29}{130}

消掉

\frac{- 12 + 16 29}{130} : \frac{2 ( 4 29 - 3 )}{65}

\frac{- 12 + 16 29}{130}

分解

- 12 + 1629 : 4 (- 3 + 429)

- 12 + 1629

改写为

= - 4 \cdot 3 + 4 \cdot 429

因式分解出通项

4

= 4 (- 3 + 429)

= \frac{4 ( - 3 + 4 29 )}{130}

约分:

2

= \frac{2 ( 4 29 - 3 )}{65}

= - \frac{2 ( 4 29 - 3 )}{65}

u = \frac{- 12 - 16 29}{2 ( - 65 )} : \frac{2 ( 3 + 4 29 )}{65}

\frac{- 12 - 16 29}{2 ( - 65 )}

去除括号:

(- a) = - a

= \frac{- 12 - 16 29}{- 2 \cdot 65}

数字相乘:

2 \cdot 65 = 130

= \frac{- 12 - 16 29}{- 130}

使用分式法则:

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

- 12 - 1629 = - (12 + 1629)

= \frac{12 + 16 29}{130}

分解

12 + 1629 : 4 (3 + 429)

12 + 1629

改写为

= 4 \cdot 3 + 4 \cdot 429

因式分解出通项

4

= 4 (3 + 429)

= \frac{4 ( 3 + 4 29 )}{130}

约分:

2

= \frac{2 ( 3 + 4 29 )}{65}

二次方程组的解是:

u = - \frac{2 ( 4 29 - 3 )}{65}, u = \frac{2 ( 3 + 4 29 )}{65}

u = cos (x)

代回

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}, cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}, cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = - \frac{2 ( 4 29 - 3 )}{65} : x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

使用反三角函数性质

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

的通解

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

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

cos (x) = \frac{2 ( 3 + 4 29 )}{65} : x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

使用反三角函数性质

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

的通解

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

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

合并所有解

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

将解代入原方程进行验证

将它们代入

8 sin (x) = cos (x) - 6

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

的解:

假

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

对于

8 sin (x) = cos (x) - 6

代入

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

8 sin (arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) = cos (arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) - 6

整理后得

6.57048 \dots = - 6.57048 \dots

\Rightarrow 假

检验

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

的解:

真

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

对于

8 sin (x) = cos (x) - 6

代入

x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

8 sin (- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) = cos (- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) - 6

整理后得

- 6.57048 \dots = - 6.57048 \dots

\Rightarrow 真

检验

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

的解:

假

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

对于

8 sin (x) = cos (x) - 6

代入

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

8 sin (arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) = cos (arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) - 6

整理后得

5.24490 \dots = - 5.24490 \dots

\Rightarrow 假

检验

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

的解:

真

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

对于

8 sin (x) = cos (x) - 6

代入

x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

8 sin (2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) = cos (2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) - 6

整理后得

- 5.24490 \dots = - 5.24490 \dots

\Rightarrow 真

x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

以小数形式表示解

x = - 2.17788 \dots + 2 πn, x = 2 π - 0.71499 \dots + 2 πn

作图

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