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受欢迎的三角函数 >

9sin(x)+6cos(x)=10

解答

9 sin (x) + 6 cos (x) = 10

解答

x = 1.37386 \dots + 2 πn, x = 0.59172 \dots + 2 πn

+1

度数

x = 78.71680 \dots^{\circ} + 36 0^{\circ} n, x = 33.90306 \dots^{\circ} + 36 0^{\circ} n

求解步骤

9 sin (x) + 6 cos (x) = 10

两边减去

6 cos (x)

9 sin (x) = 10 - 6 cos (x)

两边进行平方

(9 sin (x))^{2} = (10 - 6 cos (x))^{2}

两边减去

(10 - 6 cos (x))^{2}

81 sin^{2} (x) - 100 + 120 cos (x) - 36 cos^{2} (x) = 0

使用三角恒等式改写

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 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)

= - 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x))

化简

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x)) : 120 cos (x) - 117 cos^{2} (x) - 19

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x))

乘开

81 (1 - cos^{2} (x)) : 81 - 81 cos^{2} (x)

81 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 81, b = 1, c = cos^{2} (x)

= 81 \cdot 1 - 81 cos^{2} (x)

数字相乘:

81 \cdot 1 = 81

= 81 - 81 cos^{2} (x)

= - 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x)

化简

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x) : 120 cos (x) - 117 cos^{2} (x) - 19

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x)

对同类项分组

= 120 cos (x) - 36 cos^{2} (x) - 81 cos^{2} (x) - 100 + 81

同类项相加:

- 36 cos^{2} (x) - 81 cos^{2} (x) = - 117 cos^{2} (x)

= 120 cos (x) - 117 cos^{2} (x) - 100 + 81

数字相加/相减:

- 100 + 81 = - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

- 19 - 117 cos^{2} (x) + 120 cos (x) = 0

用替代法求解

- 19 - 117 cos^{2} (x) + 120 cos (x) = 0

令

: cos (x) = u

- 19 - 117 u^{2} + 120 u = 0

- 19 - 117 u^{2} + 120 u = 0 : u = \frac{20 - 3 17}{39}, u = \frac{20 + 3 17}{39}

- 19 - 117 u^{2} + 120 u = 0

改写成标准形式

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

- 117 u^{2} + 120 u - 19 = 0

使用求根公式求解

- 117 u^{2} + 120 u - 19 = 0

二次方程求根公式:

若

a = - 117, b = 120, c = - 19

u_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 117 ) ( - 19 )}{2 ( - 117 )}

u_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 117 ) ( - 19 )}{2 ( - 117 )}

12 0^{2} - 4 (- 117) (- 19) = 1817

12 0^{2} - 4 (- 117) (- 19)

使用法则

- (- a) = a

= 12 0^{2} - 4 \cdot 117 \cdot 19

数字相乘:

4 \cdot 117 \cdot 19 = 8892

= 12 0^{2} - 8892

12 0^{2} = 14400

= 14400 - 8892

数字相减:

14400 - 8892 = 5508

= 5508

5508

质因数分解:

2^{2} \cdot 3^{4} \cdot 17

5508

5508

除以

25508 = 2754 \cdot 2

= 2 \cdot 2754

2754

除以

22754 = 1377 \cdot 2

= 2 \cdot 2 \cdot 1377

1377

除以

31377 = 459 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 459

459

除以

3459 = 153 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 153

153

除以

3153 = 51 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 51

51

除以

351 = 17 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 17

2, 3, 17

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 17

= 2^{2} \cdot 3^{4} \cdot 17

= 3^{4} \cdot 2^{2} \cdot 17

使用根式运算法则:

n ab = n a n b

= 17 2^{2} 3^{4}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 217 3^{4}

使用根式运算法则:

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

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 3^{2} \cdot 217

整理后得

= 1817

u_{1, 2} = \frac{- 120 \pm 18 17}{2 ( - 117 )}

将解分隔开

u_{1} = \frac{- 120 + 18 17}{2 ( - 117 )}, u_{2} = \frac{- 120 - 18 17}{2 ( - 117 )}

u = \frac{- 120 + 18 17}{2 ( - 117 )} : \frac{20 - 3 17}{39}

\frac{- 120 + 18 17}{2 ( - 117 )}

去除括号:

(- a) = - a

= \frac{- 120 + 18 17}{- 2 \cdot 117}

数字相乘:

2 \cdot 117 = 234

= \frac{- 120 + 18 17}{- 234}

使用分式法则:

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

- 120 + 1817 = - (120 - 1817)

= \frac{120 - 18 17}{234}

分解

120 - 1817 : 6 (20 - 317)

120 - 1817

改写为

= 6 \cdot 20 - 6 \cdot 317

因式分解出通项

6

= 6 (20 - 317)

= \frac{6 ( 20 - 3 17 )}{234}

约分:

6

= \frac{20 - 3 17}{39}

u = \frac{- 120 - 18 17}{2 ( - 117 )} : \frac{20 + 3 17}{39}

\frac{- 120 - 18 17}{2 ( - 117 )}

去除括号:

(- a) = - a

= \frac{- 120 - 18 17}{- 2 \cdot 117}

数字相乘:

2 \cdot 117 = 234

= \frac{- 120 - 18 17}{- 234}

使用分式法则:

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

- 120 - 1817 = - (120 + 1817)

= \frac{120 + 18 17}{234}

分解

120 + 1817 : 6 (20 + 317)

120 + 1817

改写为

= 6 \cdot 20 + 6 \cdot 317

因式分解出通项

6

= 6 (20 + 317)

= \frac{6 ( 20 + 3 17 )}{234}

约分:

6

= \frac{20 + 3 17}{39}

二次方程组的解是:

u = \frac{20 - 3 17}{39}, u = \frac{20 + 3 17}{39}

u = cos (x)

代回

cos (x) = \frac{20 - 3 17}{39}, cos (x) = \frac{20 + 3 17}{39}

cos (x) = \frac{20 - 3 17}{39}, cos (x) = \frac{20 + 3 17}{39}

cos (x) = \frac{20 - 3 17}{39} : x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

cos (x) = \frac{20 - 3 17}{39}

使用反三角函数性质

cos (x) = \frac{20 - 3 17}{39}

cos (x) = \frac{20 - 3 17}{39}

的通解

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

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

cos (x) = \frac{20 + 3 17}{39} : x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

cos (x) = \frac{20 + 3 17}{39}

使用反三角函数性质

cos (x) = \frac{20 + 3 17}{39}

cos (x) = \frac{20 + 3 17}{39}

的通解

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

x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

合并所有解

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn, x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

将解代入原方程进行验证

将它们代入

9 sin (x) + 6 cos (x) = 10

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{20 - 3 17}{39}) + 2 πn

的解:

真

arccos (\frac{20 - 3 17}{39}) + 2 πn

代入

n = 1

arccos (\frac{20 - 3 17}{39}) + 2 π 1

对于

9 sin (x) + 6 cos (x) = 10

代入

x = arccos (\frac{20 - 3 17}{39}) + 2 π 1

9 sin (arccos (\frac{20 - 3 17}{39}) + 2 π 1) + 6 cos (arccos (\frac{20 - 3 17}{39}) + 2 π 1) = 10

整理后得

10 = 10

\Rightarrow 真

检验

2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

的解:

假

2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

代入

n = 1

2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1

对于

9 sin (x) + 6 cos (x) = 10

代入

x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1

9 sin (2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1) + 6 cos (2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1) = 10

整理后得

- 7.65209 \dots = 10

\Rightarrow 假

检验

arccos (\frac{20 + 3 17}{39}) + 2 πn

的解:

真

arccos (\frac{20 + 3 17}{39}) + 2 πn

代入

n = 1

arccos (\frac{20 + 3 17}{39}) + 2 π 1

对于

9 sin (x) + 6 cos (x) = 10

代入

x = arccos (\frac{20 + 3 17}{39}) + 2 π 1

9 sin (arccos (\frac{20 + 3 17}{39}) + 2 π 1) + 6 cos (arccos (\frac{20 + 3 17}{39}) + 2 π 1) = 10

整理后得

10 = 10

\Rightarrow 真

检验

2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

的解:

假

2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

代入

n = 1

2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1

对于

9 sin (x) + 6 cos (x) = 10

代入

x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1

9 sin (2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1) + 6 cos (2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1) = 10

整理后得

- 0.04021 \dots = 10

\Rightarrow 假

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = arccos (\frac{20 + 3 17}{39}) + 2 πn

以小数形式表示解

x = 1.37386 \dots + 2 πn, x = 0.59172 \dots + 2 πn

作图

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