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受欢迎的三角函数 >

8sin(x)=sqrt(41-40cos(x))

解答

8 sin (x) = 41 - 40 cos (x)

解答

x = 1.94286 \dots + 2 πn, x = 0.15153 \dots + 2 πn

+1

度数

x = 111.31781 \dots^{\circ} + 36 0^{\circ} n, x = 8.68218 \dots^{\circ} + 36 0^{\circ} n

求解步骤

8 sin (x) = 41 - 40 cos (x)

两边进行平方

(8 sin (x))^{2} = (41 - 40 cos (x))^{2}

两边减去

41 - 40 cos (x)^{2}

64 sin^{2} (x) - 41 + 40 cos (x) = 0

使用三角恒等式改写

- 41 + 40 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)

= - 41 + 40 cos (x) + 64 (1 - cos^{2} (x))

化简

- 41 + 40 cos (x) + 64 (1 - cos^{2} (x)) : 40 cos (x) - 64 cos^{2} (x) + 23

- 41 + 40 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)

= - 41 + 40 cos (x) + 64 - 64 cos^{2} (x)

化简

- 41 + 40 cos (x) + 64 - 64 cos^{2} (x) : 40 cos (x) - 64 cos^{2} (x) + 23

- 41 + 40 cos (x) + 64 - 64 cos^{2} (x)

对同类项分组

= 40 cos (x) - 64 cos^{2} (x) - 41 + 64

数字相加/相减:

- 41 + 64 = 23

= 40 cos (x) - 64 cos^{2} (x) + 23

= 40 cos (x) - 64 cos^{2} (x) + 23

= 40 cos (x) - 64 cos^{2} (x) + 23

23 + 40 cos (x) - 64 cos^{2} (x) = 0

用替代法求解

23 + 40 cos (x) - 64 cos^{2} (x) = 0

令

: cos (x) = u

23 + 40 u - 64 u^{2} = 0

23 + 40 u - 64 u^{2} = 0 : u = - \frac{- 5 + 3 13}{16}, u = \frac{5 + 3 13}{16}

23 + 40 u - 64 u^{2} = 0

改写成标准形式

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

- 64 u^{2} + 40 u + 23 = 0

使用求根公式求解

- 64 u^{2} + 40 u + 23 = 0

二次方程求根公式:

若

a = - 64, b = 40, c = 23

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 64 ) \cdot 23}{2 ( - 64 )}

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 64 ) \cdot 23}{2 ( - 64 )}

4 0^{2} - 4 (- 64) \cdot 23 = 2413

4 0^{2} - 4 (- 64) \cdot 23

使用法则

- (- a) = a

= 4 0^{2} + 4 \cdot 64 \cdot 23

数字相乘:

4 \cdot 64 \cdot 23 = 5888

= 4 0^{2} + 5888

4 0^{2} = 1600

= 1600 + 5888

数字相加:

1600 + 5888 = 7488

= 7488

7488

质因数分解:

2^{6} \cdot 3^{2} \cdot 13

7488

7488

除以

27488 = 3744 \cdot 2

= 2 \cdot 3744

3744

除以

23744 = 1872 \cdot 2

= 2 \cdot 2 \cdot 1872

1872

除以

21872 = 936 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 936

936

除以

2936 = 468 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 468

468

除以

2468 = 234 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 234

234

除以

2234 = 117 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 117

117

除以

3117 = 39 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 39

39

除以

339 = 13 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 13

2, 3, 13

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 3 \cdot 3 \cdot 13

= 2^{6} \cdot 3^{2} \cdot 13

= 2^{6} \cdot 3^{2} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{6} 3^{2}

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} 13 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{3} \cdot 313

整理后得

= 2413

u_{1, 2} = \frac{- 40 \pm 24 13}{2 ( - 64 )}

将解分隔开

u_{1} = \frac{- 40 + 24 13}{2 ( - 64 )}, u_{2} = \frac{- 40 - 24 13}{2 ( - 64 )}

u = \frac{- 40 + 24 13}{2 ( - 64 )} : - \frac{- 5 + 3 13}{16}

\frac{- 40 + 24 13}{2 ( - 64 )}

去除括号:

(- a) = - a

= \frac{- 40 + 24 13}{- 2 \cdot 64}

数字相乘:

2 \cdot 64 = 128

= \frac{- 40 + 24 13}{- 128}

使用分式法则:

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

= - \frac{- 40 + 24 13}{128}

消掉

\frac{- 40 + 24 13}{128} : \frac{3 13 - 5}{16}

\frac{- 40 + 24 13}{128}

分解

- 40 + 2413 : 8 (- 5 + 313)

- 40 + 2413

改写为

= - 8 \cdot 5 + 8 \cdot 313

因式分解出通项

8

= 8 (- 5 + 313)

= \frac{8 ( - 5 + 3 13 )}{128}

约分:

8

= \frac{- 5 + 3 13}{16}

= - \frac{3 13 - 5}{16}

= - \frac{- 5 + 3 13}{16}

u = \frac{- 40 - 24 13}{2 ( - 64 )} : \frac{5 + 3 13}{16}

\frac{- 40 - 24 13}{2 ( - 64 )}

去除括号:

(- a) = - a

= \frac{- 40 - 24 13}{- 2 \cdot 64}

数字相乘:

2 \cdot 64 = 128

= \frac{- 40 - 24 13}{- 128}

使用分式法则:

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

- 40 - 2413 = - (40 + 2413)

= \frac{40 + 24 13}{128}

分解

40 + 2413 : 8 (5 + 313)

40 + 2413

改写为

= 8 \cdot 5 + 8 \cdot 313

因式分解出通项

8

= 8 (5 + 313)

= \frac{8 ( 5 + 3 13 )}{128}

约分:

8

= \frac{5 + 3 13}{16}

二次方程组的解是:

u = - \frac{- 5 + 3 13}{16}, u = \frac{5 + 3 13}{16}

u = cos (x)

代回

cos (x) = - \frac{- 5 + 3 13}{16}, cos (x) = \frac{5 + 3 13}{16}

cos (x) = - \frac{- 5 + 3 13}{16}, cos (x) = \frac{5 + 3 13}{16}

cos (x) = - \frac{- 5 + 3 13}{16} : x = arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = - arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

cos (x) = - \frac{- 5 + 3 13}{16}

使用反三角函数性质

cos (x) = - \frac{- 5 + 3 13}{16}

cos (x) = - \frac{- 5 + 3 13}{16}

的通解

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

x = arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = - arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

x = arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = - arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

cos (x) = \frac{5 + 3 13}{16} : x = arccos (\frac{5 + 3 13}{16}) + 2 πn, x = 2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

cos (x) = \frac{5 + 3 13}{16}

使用反三角函数性质

cos (x) = \frac{5 + 3 13}{16}

cos (x) = \frac{5 + 3 13}{16}

的通解

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

x = arccos (\frac{5 + 3 13}{16}) + 2 πn, x = 2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

x = arccos (\frac{5 + 3 13}{16}) + 2 πn, x = 2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

合并所有解

x = arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = - arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = arccos (\frac{5 + 3 13}{16}) + 2 πn, x = 2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

将解代入原方程进行验证

将它们代入

8 sin (x) = 41 - 40 cos (x)

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

的解:

真

arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

代入

n = 1

arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1

对于

8 sin (x) = 41 - 40 cos (x)

代入

x = arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1

8 sin (arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1) = 41 - 40 cos (arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1)

整理后得

7.45262 \dots = 7.45262 \dots

\Rightarrow 真

检验

- arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

的解:

假

- arccos (- \frac{- 5 + 3 13}{16}) + 2 πn

代入

n = 1

- arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1

对于

8 sin (x) = 41 - 40 cos (x)

代入

x = - arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1

8 sin (- arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1) = 41 - 40 cos (- arccos (- \frac{- 5 + 3 13}{16}) + 2 π 1)

整理后得

- 7.45262 \dots = 7.45262 \dots

\Rightarrow 假

检验

arccos (\frac{5 + 3 13}{16}) + 2 πn

的解:

真

arccos (\frac{5 + 3 13}{16}) + 2 πn

代入

n = 1

arccos (\frac{5 + 3 13}{16}) + 2 π 1

对于

8 sin (x) = 41 - 40 cos (x)

代入

x = arccos (\frac{5 + 3 13}{16}) + 2 π 1

8 sin (arccos (\frac{5 + 3 13}{16}) + 2 π 1) = 41 - 40 cos (arccos (\frac{5 + 3 13}{16}) + 2 π 1)

整理后得

1.20762 \dots = 1.20762 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

的解:

假

2 π - arccos (\frac{5 + 3 13}{16}) + 2 πn

代入

n = 1

2 π - arccos (\frac{5 + 3 13}{16}) + 2 π 1

对于

8 sin (x) = 41 - 40 cos (x)

代入

x = 2 π - arccos (\frac{5 + 3 13}{16}) + 2 π 1

8 sin (2 π - arccos (\frac{5 + 3 13}{16}) + 2 π 1) = 41 - 40 cos (2 π - arccos (\frac{5 + 3 13}{16}) + 2 π 1)

整理后得

- 1.20762 \dots = 1.20762 \dots

\Rightarrow 假

x = arccos (- \frac{- 5 + 3 13}{16}) + 2 πn, x = arccos (\frac{5 + 3 13}{16}) + 2 πn

以小数形式表示解

x = 1.94286 \dots + 2 πn, x = 0.15153 \dots + 2 πn

作图

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