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受欢迎的三角函数 >

74=58cos(x)+71sin(x)

解答

74 = 58 cos (x) + 71 sin (x)

解答

x = 0.25437 \dots + 2 πn, x = 1.51729 \dots + 2 πn

+1

度数

x = 14.57452 \dots^{\circ} + 36 0^{\circ} n, x = 86.93460 \dots^{\circ} + 36 0^{\circ} n

求解步骤

74 = 58 cos (x) + 71 sin (x)

两边减去

71 sin (x)

58 cos (x) = 74 - 71 sin (x)

两边进行平方

(58 cos (x))^{2} = (74 - 71 sin (x))^{2}

两边减去

(74 - 71 sin (x))^{2}

3364 cos^{2} (x) - 5476 + 10508 sin (x) - 5041 sin^{2} (x) = 0

使用三角恒等式改写

- 5476 + 10508 sin (x) + 3364 cos^{2} (x) - 5041 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 5476 + 10508 sin (x) + 3364 (1 - sin^{2} (x)) - 5041 sin^{2} (x)

化简

- 5476 + 10508 sin (x) + 3364 (1 - sin^{2} (x)) - 5041 sin^{2} (x) : 10508 sin (x) - 8405 sin^{2} (x) - 2112

- 5476 + 10508 sin (x) + 3364 (1 - sin^{2} (x)) - 5041 sin^{2} (x)

乘开

3364 (1 - sin^{2} (x)) : 3364 - 3364 sin^{2} (x)

3364 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3364, b = 1, c = sin^{2} (x)

= 3364 \cdot 1 - 3364 sin^{2} (x)

数字相乘:

3364 \cdot 1 = 3364

= 3364 - 3364 sin^{2} (x)

= - 5476 + 10508 sin (x) + 3364 - 3364 sin^{2} (x) - 5041 sin^{2} (x)

化简

- 5476 + 10508 sin (x) + 3364 - 3364 sin^{2} (x) - 5041 sin^{2} (x) : 10508 sin (x) - 8405 sin^{2} (x) - 2112

- 5476 + 10508 sin (x) + 3364 - 3364 sin^{2} (x) - 5041 sin^{2} (x)

同类项相加:

- 3364 sin^{2} (x) - 5041 sin^{2} (x) = - 8405 sin^{2} (x)

= - 5476 + 10508 sin (x) + 3364 - 8405 sin^{2} (x)

对同类项分组

= 10508 sin (x) - 8405 sin^{2} (x) - 5476 + 3364

数字相加/相减:

- 5476 + 3364 = - 2112

= 10508 sin (x) - 8405 sin^{2} (x) - 2112

= 10508 sin (x) - 8405 sin^{2} (x) - 2112

= 10508 sin (x) - 8405 sin^{2} (x) - 2112

- 2112 + 10508 sin (x) - 8405 sin^{2} (x) = 0

用替代法求解

- 2112 + 10508 sin (x) - 8405 sin^{2} (x) = 0

令

: sin (x) = u

- 2112 + 10508 u - 8405 u^{2} = 0

- 2112 + 10508 u - 8405 u^{2} = 0 : u = \frac{2 ( 2627 - 29 2929 )}{8405}, u = \frac{2 ( 2627 + 29 2929 )}{8405}

- 2112 + 10508 u - 8405 u^{2} = 0

改写成标准形式

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

- 8405 u^{2} + 10508 u - 2112 = 0

使用求根公式求解

- 8405 u^{2} + 10508 u - 2112 = 0

二次方程求根公式:

若

a = - 8405, b = 10508, c = - 2112

u_{1, 2} = \frac{- 10508 \pm 1050 8 ^{2} - 4 ( - 8405 ) ( - 2112 )}{2 ( - 8405 )}

u_{1, 2} = \frac{- 10508 \pm 1050 8 ^{2} - 4 ( - 8405 ) ( - 2112 )}{2 ( - 8405 )}

1050 8^{2} - 4 (- 8405) (- 2112) = 1162929

1050 8^{2} - 4 (- 8405) (- 2112)

使用法则

- (- a) = a

= 1050 8^{2} - 4 \cdot 8405 \cdot 2112

数字相乘:

4 \cdot 8405 \cdot 2112 = 71005440

= 1050 8^{2} - 71005440

1050 8^{2} = 110418064

= 110418064 - 71005440

数字相减:

110418064 - 71005440 = 39412624

= 39412624

39412624

质因数分解:

2^{4} \cdot 2 9^{3} \cdot 101

39412624

= 2^{4} \cdot 2 9^{3} \cdot 101

使用指数法则:

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

= 2^{4} \cdot 2 9^{2} \cdot 29 \cdot 101

使用根式运算法则:

n ab = n a n b

= 2^{4} 2 9^{2} 29 \cdot 101

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} 2 9^{2} 29 \cdot 101

使用根式运算法则:

n a^{n} = a

2 9^{2} = 29

= 2^{2} \cdot 29 29 \cdot 101

整理后得

= 1162929

u_{1, 2} = \frac{- 10508 \pm 116 2929}{2 ( - 8405 )}

将解分隔开

u_{1} = \frac{- 10508 + 116 2929}{2 ( - 8405 )}, u_{2} = \frac{- 10508 - 116 2929}{2 ( - 8405 )}

u = \frac{- 10508 + 116 2929}{2 ( - 8405 )} : \frac{2 ( 2627 - 29 2929 )}{8405}

\frac{- 10508 + 116 2929}{2 ( - 8405 )}

去除括号:

(- a) = - a

= \frac{- 10508 + 116 2929}{- 2 \cdot 8405}

数字相乘:

2 \cdot 8405 = 16810

= \frac{- 10508 + 116 2929}{- 16810}

使用分式法则:

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

- 10508 + 1162929 = - (10508 - 1162929)

= \frac{10508 - 116 2929}{16810}

分解

10508 - 1162929 : 4 (2627 - 292929)

10508 - 1162929

改写为

= 4 \cdot 2627 - 4 \cdot 292929

因式分解出通项

4

= 4 (2627 - 292929)

= \frac{4 ( 2627 - 29 2929 )}{16810}

约分:

2

= \frac{2 ( 2627 - 29 2929 )}{8405}

u = \frac{- 10508 - 116 2929}{2 ( - 8405 )} : \frac{2 ( 2627 + 29 2929 )}{8405}

\frac{- 10508 - 116 2929}{2 ( - 8405 )}

去除括号:

(- a) = - a

= \frac{- 10508 - 116 2929}{- 2 \cdot 8405}

数字相乘:

2 \cdot 8405 = 16810

= \frac{- 10508 - 116 2929}{- 16810}

使用分式法则:

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

- 10508 - 1162929 = - (10508 + 1162929)

= \frac{10508 + 116 2929}{16810}

分解

10508 + 1162929 : 4 (2627 + 292929)

10508 + 1162929

改写为

= 4 \cdot 2627 + 4 \cdot 292929

因式分解出通项

4

= 4 (2627 + 292929)

= \frac{4 ( 2627 + 29 2929 )}{16810}

约分:

2

= \frac{2 ( 2627 + 29 2929 )}{8405}

二次方程组的解是:

u = \frac{2 ( 2627 - 29 2929 )}{8405}, u = \frac{2 ( 2627 + 29 2929 )}{8405}

u = sin (x)

代回

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405}, sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405}

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405}, sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405}

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405} : x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405}

使用反三角函数性质

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405}

sin (x) = \frac{2 ( 2627 - 29 2929 )}{8405}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405} : x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405}

使用反三角函数性质

sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405}

sin (x) = \frac{2 ( 2627 + 29 2929 )}{8405}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

合并所有解

x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn, x = π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

将解代入原方程进行验证

将它们代入

58 cos (x) + 71 sin (x) = 74

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

的解:

真

arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

代入

n = 1

arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1

对于

58 cos (x) + 71 sin (x) = 74

代入

x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1

58 cos (arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1) + 71 sin (arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1) = 74

整理后得

74 = 74

\Rightarrow 真

检验

π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

的解:

假

π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn

代入

n = 1

π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1

对于

58 cos (x) + 71 sin (x) = 74

代入

x = π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1

58 cos (π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1) + 71 sin (π - arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 π 1) = 74

整理后得

- 38.26725 \dots = 74

\Rightarrow 假

检验

arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

的解:

真

arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

代入

n = 1

arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1

对于

58 cos (x) + 71 sin (x) = 74

代入

x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1

58 cos (arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1) + 71 sin (arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1) = 74

整理后得

74 = 74

\Rightarrow 真

检验

π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

的解:

假

π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

代入

n = 1

π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1

对于

58 cos (x) + 71 sin (x) = 74

代入

x = π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1

58 cos (π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1) + 71 sin (π - arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 π 1) = 74

整理后得

67.79681 \dots = 74

\Rightarrow 假

x = arcsin (\frac{2 ( 2627 - 29 2929 )}{8405}) + 2 πn, x = arcsin (\frac{2 ( 2627 + 29 2929 )}{8405}) + 2 πn

以小数形式表示解

x = 0.25437 \dots + 2 πn, x = 1.51729 \dots + 2 πn

作图

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