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受欢迎的三角函数 >

7sin(x)=cos(x)-4

解答

7 sin (x) = cos (x) - 4

解答

x = - 2.39843 \dots + 2 πn, x = 2 π - 0.45936 \dots + 2 πn

+1

度数

x = - 137.41999 \dots^{\circ} + 36 0^{\circ} n, x = 333.68020 \dots^{\circ} + 36 0^{\circ} n

求解步骤

7 sin (x) = cos (x) - 4

两边进行平方

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

两边减去

(cos (x) - 4)^{2}

49 sin^{2} (x) - cos^{2} (x) + 8 cos (x) - 16 = 0

使用三角恒等式改写

- 16 - cos^{2} (x) + 49 sin^{2} (x) + 8 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)

= - 16 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 8 cos (x)

化简

- 16 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 50 cos^{2} (x) + 33

- 16 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 8 cos (x)

乘开

49 (1 - cos^{2} (x)) : 49 - 49 cos^{2} (x)

49 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 49, b = 1, c = cos^{2} (x)

= 49 \cdot 1 - 49 cos^{2} (x)

数字相乘:

49 \cdot 1 = 49

= 49 - 49 cos^{2} (x)

= - 16 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 8 cos (x)

化简

- 16 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 8 cos (x) : 8 cos (x) - 50 cos^{2} (x) + 33

- 16 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 8 cos (x)

对同类项分组

= - cos^{2} (x) - 49 cos^{2} (x) + 8 cos (x) - 16 + 49

同类项相加:

- cos^{2} (x) - 49 cos^{2} (x) = - 50 cos^{2} (x)

= - 50 cos^{2} (x) + 8 cos (x) - 16 + 49

数字相加/相减:

- 16 + 49 = 33

= 8 cos (x) - 50 cos^{2} (x) + 33

= 8 cos (x) - 50 cos^{2} (x) + 33

= 8 cos (x) - 50 cos^{2} (x) + 33

33 - 50 cos^{2} (x) + 8 cos (x) = 0

用替代法求解

33 - 50 cos^{2} (x) + 8 cos (x) = 0

令

: cos (x) = u

33 - 50 u^{2} + 8 u = 0

33 - 50 u^{2} + 8 u = 0 : u = - \frac{- 4 + 7 34}{50}, u = \frac{4 + 7 34}{50}

33 - 50 u^{2} + 8 u = 0

改写成标准形式

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

- 50 u^{2} + 8 u + 33 = 0

使用求根公式求解

- 50 u^{2} + 8 u + 33 = 0

二次方程求根公式:

若

a = - 50, b = 8, c = 33

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 50 ) \cdot 33}{2 ( - 50 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 50 ) \cdot 33}{2 ( - 50 )}

8^{2} - 4 (- 50) \cdot 33 = 1434

8^{2} - 4 (- 50) \cdot 33

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 50 \cdot 33

数字相乘:

4 \cdot 50 \cdot 33 = 6600

= 8^{2} + 6600

8^{2} = 64

= 64 + 6600

数字相加:

64 + 6600 = 6664

= 6664

6664

质因数分解:

2^{3} \cdot 7^{2} \cdot 17

6664

6664

除以

26664 = 3332 \cdot 2

= 2 \cdot 3332

3332

除以

23332 = 1666 \cdot 2

= 2 \cdot 2 \cdot 1666

1666

除以

21666 = 833 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 833

833

除以

7833 = 119 \cdot 7

= 2 \cdot 2 \cdot 2 \cdot 7 \cdot 119

119

除以

7119 = 17 \cdot 7

= 2 \cdot 2 \cdot 2 \cdot 7 \cdot 7 \cdot 17

2, 7, 17

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 7 \cdot 7 \cdot 17

= 2^{3} \cdot 7^{2} \cdot 17

= 2^{3} \cdot 7^{2} \cdot 17

使用指数法则:

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

= 2^{2} \cdot 7^{2} \cdot 2 \cdot 17

使用根式运算法则:

n ab = n a n b

= 2^{2} 7^{2} 2 \cdot 17

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 7^{2} 2 \cdot 17

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 2 \cdot 7 2 \cdot 17

整理后得

= 1434

u_{1, 2} = \frac{- 8 \pm 14 34}{2 ( - 50 )}

将解分隔开

u_{1} = \frac{- 8 + 14 34}{2 ( - 50 )}, u_{2} = \frac{- 8 - 14 34}{2 ( - 50 )}

u = \frac{- 8 + 14 34}{2 ( - 50 )} : - \frac{- 4 + 7 34}{50}

\frac{- 8 + 14 34}{2 ( - 50 )}

去除括号:

(- a) = - a

= \frac{- 8 + 14 34}{- 2 \cdot 50}

数字相乘:

2 \cdot 50 = 100

= \frac{- 8 + 14 34}{- 100}

使用分式法则:

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

= - \frac{- 8 + 14 34}{100}

消掉

\frac{- 8 + 14 34}{100} : \frac{7 34 - 4}{50}

\frac{- 8 + 14 34}{100}

分解

- 8 + 1434 : 2 (- 4 + 734)

- 8 + 1434

改写为

= - 2 \cdot 4 + 2 \cdot 734

因式分解出通项

2

= 2 (- 4 + 734)

= \frac{2 ( - 4 + 7 34 )}{100}

约分:

2

= \frac{- 4 + 7 34}{50}

= - \frac{7 34 - 4}{50}

= - \frac{- 4 + 7 34}{50}

u = \frac{- 8 - 14 34}{2 ( - 50 )} : \frac{4 + 7 34}{50}

\frac{- 8 - 14 34}{2 ( - 50 )}

去除括号:

(- a) = - a

= \frac{- 8 - 14 34}{- 2 \cdot 50}

数字相乘:

2 \cdot 50 = 100

= \frac{- 8 - 14 34}{- 100}

使用分式法则:

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

- 8 - 1434 = - (8 + 1434)

= \frac{8 + 14 34}{100}

分解

8 + 1434 : 2 (4 + 734)

8 + 1434

改写为

= 2 \cdot 4 + 2 \cdot 734

因式分解出通项

2

= 2 (4 + 734)

= \frac{2 ( 4 + 7 34 )}{100}

约分:

2

= \frac{4 + 7 34}{50}

二次方程组的解是:

u = - \frac{- 4 + 7 34}{50}, u = \frac{4 + 7 34}{50}

u = cos (x)

代回

cos (x) = - \frac{- 4 + 7 34}{50}, cos (x) = \frac{4 + 7 34}{50}

cos (x) = - \frac{- 4 + 7 34}{50}, cos (x) = \frac{4 + 7 34}{50}

cos (x) = - \frac{- 4 + 7 34}{50} : x = arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

cos (x) = - \frac{- 4 + 7 34}{50}

使用反三角函数性质

cos (x) = - \frac{- 4 + 7 34}{50}

cos (x) = - \frac{- 4 + 7 34}{50}

的通解

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

x = arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

x = arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

cos (x) = \frac{4 + 7 34}{50} : x = arccos (\frac{4 + 7 34}{50}) + 2 πn, x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

cos (x) = \frac{4 + 7 34}{50}

使用反三角函数性质

cos (x) = \frac{4 + 7 34}{50}

cos (x) = \frac{4 + 7 34}{50}

的通解

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

x = arccos (\frac{4 + 7 34}{50}) + 2 πn, x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

x = arccos (\frac{4 + 7 34}{50}) + 2 πn, x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

合并所有解

x = arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = arccos (\frac{4 + 7 34}{50}) + 2 πn, x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

将解代入原方程进行验证

将它们代入

7 sin (x) = cos (x) - 4

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

的解:

假

arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

代入

n = 1

arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 4

代入

x = arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1

7 sin (arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1) = cos (arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1) - 4

整理后得

4.73633 \dots = - 4.73633 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

的解:

真

- arccos (- \frac{- 4 + 7 34}{50}) + 2 πn

代入

n = 1

- arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 4

代入

x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1

7 sin (- arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1) = cos (- arccos (- \frac{- 4 + 7 34}{50}) + 2 π 1) - 4

整理后得

- 4.73633 \dots = - 4.73633 \dots

\Rightarrow 真

检验

arccos (\frac{4 + 7 34}{50}) + 2 πn

的解:

假

arccos (\frac{4 + 7 34}{50}) + 2 πn

代入

n = 1

arccos (\frac{4 + 7 34}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 4

代入

x = arccos (\frac{4 + 7 34}{50}) + 2 π 1

7 sin (arccos (\frac{4 + 7 34}{50}) + 2 π 1) = cos (arccos (\frac{4 + 7 34}{50}) + 2 π 1) - 4

整理后得

3.10366 \dots = - 3.10366 \dots

\Rightarrow 假

检验

2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

的解:

真

2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

代入

n = 1

2 π - arccos (\frac{4 + 7 34}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 4

代入

x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 π 1

7 sin (2 π - arccos (\frac{4 + 7 34}{50}) + 2 π 1) = cos (2 π - arccos (\frac{4 + 7 34}{50}) + 2 π 1) - 4

整理后得

- 3.10366 \dots = - 3.10366 \dots

\Rightarrow 真

x = - arccos (- \frac{- 4 + 7 34}{50}) + 2 πn, x = 2 π - arccos (\frac{4 + 7 34}{50}) + 2 πn

以小数形式表示解

x = - 2.39843 \dots + 2 πn, x = 2 π - 0.45936 \dots + 2 πn

作图

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