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受欢迎的三角函数 >

4sin(x)+5cos(x)=6

解答

4 sin (x) + 5 cos (x) = 6

解答

x = 1.03147 \dots + 2 πn, x = 0.31800 \dots + 2 πn

+1

度数

x = 59.09912 \dots^{\circ} + 36 0^{\circ} n, x = 18.22049 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 sin (x) + 5 cos (x) = 6

两边减去

5 cos (x)

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

两边进行平方

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

两边减去

(6 - 5 cos (x))^{2}

16 sin^{2} (x) - 36 + 60 cos (x) - 25 cos^{2} (x) = 0

使用三角恒等式改写

- 36 + 16 sin^{2} (x) - 25 cos^{2} (x) + 60 cos (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 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x)

化简

- 36 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x) : 60 cos (x) - 41 cos^{2} (x) - 20

- 36 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x)

乘开

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

数字相乘:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

= - 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x)

化简

- 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x) : 60 cos (x) - 41 cos^{2} (x) - 20

- 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x)

同类项相加:

- 16 cos^{2} (x) - 25 cos^{2} (x) = - 41 cos^{2} (x)

= - 36 + 16 - 41 cos^{2} (x) + 60 cos (x)

数字相加/相减:

- 36 + 16 = - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

- 20 - 41 cos^{2} (x) + 60 cos (x) = 0

用替代法求解

- 20 - 41 cos^{2} (x) + 60 cos (x) = 0

令

: cos (x) = u

- 20 - 41 u^{2} + 60 u = 0

- 20 - 41 u^{2} + 60 u = 0 : u = \frac{2 ( 15 - 2 5 )}{41}, u = \frac{2 ( 15 + 2 5 )}{41}

- 20 - 41 u^{2} + 60 u = 0

改写成标准形式

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

- 41 u^{2} + 60 u - 20 = 0

使用求根公式求解

- 41 u^{2} + 60 u - 20 = 0

二次方程求根公式:

若

a = - 41, b = 60, c = - 20

u_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 41 ) ( - 20 )}{2 ( - 41 )}

u_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 41 ) ( - 20 )}{2 ( - 41 )}

6 0^{2} - 4 (- 41) (- 20) = 85

6 0^{2} - 4 (- 41) (- 20)

使用法则

- (- a) = a

= 6 0^{2} - 4 \cdot 41 \cdot 20

数字相乘:

4 \cdot 41 \cdot 20 = 3280

= 6 0^{2} - 3280

6 0^{2} = 3600

= 3600 - 3280

数字相减:

3600 - 3280 = 320

= 320

320

质因数分解:

2^{6} \cdot 5

320

320

除以

2320 = 160 \cdot 2

= 2 \cdot 160

160

除以

2160 = 80 \cdot 2

= 2 \cdot 2 \cdot 80

80

除以

280 = 40 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 40

40

除以

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

2, 5

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 5

= 2^{6} \cdot 5

= 2^{6} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{6}

使用根式运算法则:

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

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 5

整理后得

= 85

u_{1, 2} = \frac{- 60 \pm 8 5}{2 ( - 41 )}

将解分隔开

u_{1} = \frac{- 60 + 8 5}{2 ( - 41 )}, u_{2} = \frac{- 60 - 8 5}{2 ( - 41 )}

u = \frac{- 60 + 8 5}{2 ( - 41 )} : \frac{2 ( 15 - 2 5 )}{41}

\frac{- 60 + 8 5}{2 ( - 41 )}

去除括号:

(- a) = - a

= \frac{- 60 + 8 5}{- 2 \cdot 41}

数字相乘:

2 \cdot 41 = 82

= \frac{- 60 + 8 5}{- 82}

使用分式法则:

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

- 60 + 85 = - (60 - 85)

= \frac{60 - 8 5}{82}

分解

60 - 85 : 4 (15 - 25)

60 - 85

改写为

= 4 \cdot 15 - 4 \cdot 25

因式分解出通项

4

= 4 (15 - 25)

= \frac{4 ( 15 - 2 5 )}{82}

约分:

2

= \frac{2 ( 15 - 2 5 )}{41}

u = \frac{- 60 - 8 5}{2 ( - 41 )} : \frac{2 ( 15 + 2 5 )}{41}

\frac{- 60 - 8 5}{2 ( - 41 )}

去除括号:

(- a) = - a

= \frac{- 60 - 8 5}{- 2 \cdot 41}

数字相乘:

2 \cdot 41 = 82

= \frac{- 60 - 8 5}{- 82}

使用分式法则:

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

- 60 - 85 = - (60 + 85)

= \frac{60 + 8 5}{82}

分解

60 + 85 : 4 (15 + 25)

60 + 85

改写为

= 4 \cdot 15 + 4 \cdot 25

因式分解出通项

4

= 4 (15 + 25)

= \frac{4 ( 15 + 2 5 )}{82}

约分:

2

= \frac{2 ( 15 + 2 5 )}{41}

二次方程组的解是:

u = \frac{2 ( 15 - 2 5 )}{41}, u = \frac{2 ( 15 + 2 5 )}{41}

u = cos (x)

代回

cos (x) = \frac{2 ( 15 - 2 5 )}{41}, cos (x) = \frac{2 ( 15 + 2 5 )}{41}

cos (x) = \frac{2 ( 15 - 2 5 )}{41}, cos (x) = \frac{2 ( 15 + 2 5 )}{41}

cos (x) = \frac{2 ( 15 - 2 5 )}{41} : x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

使用反三角函数性质

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

的通解

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

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 + 2 5 )}{41} : x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

使用反三角函数性质

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

的通解

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

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

合并所有解

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

将解代入原方程进行验证

将它们代入

4 sin (x) + 5 cos (x) = 6

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

的解:

真

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

对于

4 sin (x) + 5 cos (x) = 6

代入

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

4 sin (arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

对于

4 sin (x) + 5 cos (x) = 6

代入

x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

4 sin (2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) = 6

整理后得

- 0.86445 \dots = 6

\Rightarrow 假

检验

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

的解:

真

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

对于

4 sin (x) + 5 cos (x) = 6

代入

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

4 sin (arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

对于

4 sin (x) + 5 cos (x) = 6

代入

x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

4 sin (2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) = 6

整理后得

3.49860 \dots = 6

\Rightarrow 假

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

以小数形式表示解

x = 1.03147 \dots + 2 πn, x = 0.31800 \dots + 2 πn

作图

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