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受欢迎的三角函数 >

-6sin(x)-5cos(x)=2

解答

- 6 sin (x) - 5 cos (x) = 2

解答

x = 2.70581 \dots + 2 πn, x = 2 π - 0.95369 \dots + 2 πn

+1

度数

x = 155.03164 \dots^{\circ} + 36 0^{\circ} n, x = 305.35720 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 6 sin (x) - 5 cos (x) = 2

两边加上

5 cos (x)

- 6 sin (x) = 2 + 5 cos (x)

两边进行平方

(- 6 sin (x))^{2} = (2 + 5 cos (x))^{2}

两边减去

(2 + 5 cos (x))^{2}

36 sin^{2} (x) - 4 - 20 cos (x) - 25 cos^{2} (x) = 0

使用三角恒等式改写

- 4 - 20 cos (x) - 25 cos^{2} (x) + 36 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)

= - 4 - 20 cos (x) - 25 cos^{2} (x) + 36 (1 - cos^{2} (x))

化简

- 4 - 20 cos (x) - 25 cos^{2} (x) + 36 (1 - cos^{2} (x)) : - 61 cos^{2} (x) - 20 cos (x) + 32

- 4 - 20 cos (x) - 25 cos^{2} (x) + 36 (1 - cos^{2} (x))

乘开

36 (1 - cos^{2} (x)) : 36 - 36 cos^{2} (x)

36 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 36, b = 1, c = cos^{2} (x)

= 36 \cdot 1 - 36 cos^{2} (x)

数字相乘:

36 \cdot 1 = 36

= 36 - 36 cos^{2} (x)

= - 4 - 20 cos (x) - 25 cos^{2} (x) + 36 - 36 cos^{2} (x)

化简

- 4 - 20 cos (x) - 25 cos^{2} (x) + 36 - 36 cos^{2} (x) : - 61 cos^{2} (x) - 20 cos (x) + 32

- 4 - 20 cos (x) - 25 cos^{2} (x) + 36 - 36 cos^{2} (x)

对同类项分组

= - 20 cos (x) - 25 cos^{2} (x) - 36 cos^{2} (x) - 4 + 36

同类项相加:

- 25 cos^{2} (x) - 36 cos^{2} (x) = - 61 cos^{2} (x)

= - 20 cos (x) - 61 cos^{2} (x) - 4 + 36

数字相加/相减:

- 4 + 36 = 32

= - 61 cos^{2} (x) - 20 cos (x) + 32

= - 61 cos^{2} (x) - 20 cos (x) + 32

= - 61 cos^{2} (x) - 20 cos (x) + 32

32 - 20 cos (x) - 61 cos^{2} (x) = 0

用替代法求解

32 - 20 cos (x) - 61 cos^{2} (x) = 0

令

: cos (x) = u

32 - 20 u - 61 u^{2} = 0

32 - 20 u - 61 u^{2} = 0 : u = - \frac{2 ( 5 + 3 57 )}{61}, u = \frac{2 ( 3 57 - 5 )}{61}

32 - 20 u - 61 u^{2} = 0

改写成标准形式

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

- 61 u^{2} - 20 u + 32 = 0

使用求根公式求解

- 61 u^{2} - 20 u + 32 = 0

二次方程求根公式:

若

a = - 61, b = - 20, c = 32

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 61 ) \cdot 32}{2 ( - 61 )}

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 61 ) \cdot 32}{2 ( - 61 )}

(- 20)^{2} - 4 (- 61) \cdot 32 = 1257

(- 20)^{2} - 4 (- 61) \cdot 32

使用法则

- (- a) = a

= (- 20)^{2} + 4 \cdot 61 \cdot 32

使用指数法则:

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

若

n

是偶数

(- 20)^{2} = 2 0^{2}

= 2 0^{2} + 4 \cdot 61 \cdot 32

数字相乘:

4 \cdot 61 \cdot 32 = 7808

= 2 0^{2} + 7808

2 0^{2} = 400

= 400 + 7808

数字相加:

400 + 7808 = 8208

= 8208

8208

质因数分解:

2^{4} \cdot 3^{3} \cdot 19

8208

8208

除以

28208 = 4104 \cdot 2

= 2 \cdot 4104

4104

除以

24104 = 2052 \cdot 2

= 2 \cdot 2 \cdot 2052

2052

除以

22052 = 1026 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 1026

1026

除以

21026 = 513 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 513

513

除以

3513 = 171 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 171

171

除以

3171 = 57 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 57

57

除以

357 = 19 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 19

2, 3, 19

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 19

= 2^{4} \cdot 3^{3} \cdot 19

= 2^{4} \cdot 3^{3} \cdot 19

使用指数法则:

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

= 2^{4} \cdot 3^{2} \cdot 3 \cdot 19

使用根式运算法则:

n ab = n a n b

= 2^{4} 3^{2} 3 \cdot 19

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} 3^{2} 3 \cdot 19

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{2} \cdot 3 3 \cdot 19

整理后得

= 1257

u_{1, 2} = \frac{- ( - 20 ) \pm 12 57}{2 ( - 61 )}

将解分隔开

u_{1} = \frac{- ( - 20 ) + 12 57}{2 ( - 61 )}, u_{2} = \frac{- ( - 20 ) - 12 57}{2 ( - 61 )}

u = \frac{- ( - 20 ) + 12 57}{2 ( - 61 )} : - \frac{2 ( 5 + 3 57 )}{61}

\frac{- ( - 20 ) + 12 57}{2 ( - 61 )}

去除括号:

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

= \frac{20 + 12 57}{- 2 \cdot 61}

数字相乘:

2 \cdot 61 = 122

= \frac{20 + 12 57}{- 122}

使用分式法则:

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

= - \frac{20 + 12 57}{122}

消掉

\frac{20 + 12 57}{122} : \frac{2 ( 5 + 3 57 )}{61}

\frac{20 + 12 57}{122}

分解

20 + 1257 : 4 (5 + 357)

20 + 1257

改写为

= 4 \cdot 5 + 4 \cdot 357

因式分解出通项

4

= 4 (5 + 357)

= \frac{4 ( 5 + 3 57 )}{122}

约分:

2

= \frac{2 ( 5 + 3 57 )}{61}

= - \frac{2 ( 5 + 3 57 )}{61}

u = \frac{- ( - 20 ) - 12 57}{2 ( - 61 )} : \frac{2 ( 3 57 - 5 )}{61}

\frac{- ( - 20 ) - 12 57}{2 ( - 61 )}

去除括号:

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

= \frac{20 - 12 57}{- 2 \cdot 61}

数字相乘:

2 \cdot 61 = 122

= \frac{20 - 12 57}{- 122}

使用分式法则:

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

20 - 1257 = - (1257 - 20)

= \frac{12 57 - 20}{122}

分解

1257 - 20 : 4 (357 - 5)

1257 - 20

改写为

= 4 \cdot 357 - 4 \cdot 5

因式分解出通项

4

= 4 (357 - 5)

= \frac{4 ( 3 57 - 5 )}{122}

约分:

2

= \frac{2 ( 3 57 - 5 )}{61}

二次方程组的解是:

u = - \frac{2 ( 5 + 3 57 )}{61}, u = \frac{2 ( 3 57 - 5 )}{61}

u = cos (x)

代回

cos (x) = - \frac{2 ( 5 + 3 57 )}{61}, cos (x) = \frac{2 ( 3 57 - 5 )}{61}

cos (x) = - \frac{2 ( 5 + 3 57 )}{61}, cos (x) = \frac{2 ( 3 57 - 5 )}{61}

cos (x) = - \frac{2 ( 5 + 3 57 )}{61} : x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = - arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

cos (x) = - \frac{2 ( 5 + 3 57 )}{61}

使用反三角函数性质

cos (x) = - \frac{2 ( 5 + 3 57 )}{61}

cos (x) = - \frac{2 ( 5 + 3 57 )}{61}

的通解

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

x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = - arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = - arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

cos (x) = \frac{2 ( 3 57 - 5 )}{61} : x = arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

cos (x) = \frac{2 ( 3 57 - 5 )}{61}

使用反三角函数性质

cos (x) = \frac{2 ( 3 57 - 5 )}{61}

cos (x) = \frac{2 ( 3 57 - 5 )}{61}

的通解

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

x = arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

x = arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

合并所有解

x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = - arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

将解代入原方程进行验证

将它们代入

- 6 sin (x) - 5 cos (x) = 2

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

的解:

真

arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1

对于

- 6 sin (x) - 5 cos (x) = 2

代入

x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1

- 6 sin (arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1) - 5 cos (arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

- arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

的解:

假

- arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1

对于

- 6 sin (x) - 5 cos (x) = 2

代入

x = - arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1

- 6 sin (- arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1) - 5 cos (- arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 π 1) = 2

整理后得

7.06541 \dots = 2

\Rightarrow 假

检验

arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

的解:

假

arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1

对于

- 6 sin (x) - 5 cos (x) = 2

代入

x = arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1

- 6 sin (arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1) - 5 cos (arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1) = 2

整理后得

- 7.78672 \dots = 2

\Rightarrow 假

检验

2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

的解:

真

2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1

对于

- 6 sin (x) - 5 cos (x) = 2

代入

x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1

- 6 sin (2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1) - 5 cos (2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

x = arccos (- \frac{2 ( 5 + 3 57 )}{61}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 57 - 5 )}{61}) + 2 πn

以小数形式表示解

x = 2.70581 \dots + 2 πn, x = 2 π - 0.95369 \dots + 2 πn

作图

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