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受欢迎的三角函数 >

5cos(x)+4-2sin(x)=0

解答

5 \cdot cos (x) + 4 - 2 \cdot sin (x) = 0

解答

x = 2.02750 \dots + 2 πn, x = - 2.78851 \dots + 2 πn

+1

度数

x = 116.16747 \dots^{\circ} + 36 0^{\circ} n, x = - 159.77029 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 cos (x) + 4 - 2 sin (x) = 0

两边加上

2 sin (x)

5 cos (x) + 4 = 2 sin (x)

两边进行平方

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

两边减去

(2 sin (x))^{2}

(5 cos (x) + 4)^{2} - 4 sin^{2} (x) = 0

使用三角恒等式改写

(4 + 5 cos (x))^{2} - 4 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)

= (4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x))

化简

(4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x)) : 29 cos^{2} (x) + 40 cos (x) + 12

(4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x))

(4 + 5 cos (x))^{2} : 16 + 40 cos (x) + 25 cos^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 4, b = 5 cos (x)

= 4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2}

化简

4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2} : 16 + 40 cos (x) + 25 cos^{2} (x)

4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

2 \cdot 4 \cdot 5 cos (x) = 40 cos (x)

2 \cdot 4 \cdot 5 cos (x)

数字相乘:

2 \cdot 4 \cdot 5 = 40

= 40 cos (x)

(5 cos (x))^{2} = 25 cos^{2} (x)

(5 cos (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 5^{2} cos^{2} (x)

5^{2} = 25

= 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x) - 4 (1 - cos^{2} (x))

乘开

- 4 (1 - cos^{2} (x)) : - 4 + 4 cos^{2} (x)

- 4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = cos^{2} (x)

= - 4 \cdot 1 - (- 4) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= 16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x)

化简

16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x) : 29 cos^{2} (x) + 40 cos (x) + 12

16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x)

对同类项分组

= 40 cos (x) + 25 cos^{2} (x) + 4 cos^{2} (x) + 16 - 4

同类项相加:

25 cos^{2} (x) + 4 cos^{2} (x) = 29 cos^{2} (x)

= 40 cos (x) + 29 cos^{2} (x) + 16 - 4

数字相加/相减:

16 - 4 = 12

= 29 cos^{2} (x) + 40 cos (x) + 12

= 29 cos^{2} (x) + 40 cos (x) + 12

= 29 cos^{2} (x) + 40 cos (x) + 12

12 + 29 cos^{2} (x) + 40 cos (x) = 0

用替代法求解

12 + 29 cos^{2} (x) + 40 cos (x) = 0

令

: cos (x) = u

12 + 29 u^{2} + 40 u = 0

12 + 29 u^{2} + 40 u = 0 : u = \frac{2 ( 13 - 10 )}{29}, u = - \frac{2 ( 10 + 13 )}{29}

12 + 29 u^{2} + 40 u = 0

改写成标准形式

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

29 u^{2} + 40 u + 12 = 0

使用求根公式求解

29 u^{2} + 40 u + 12 = 0

二次方程求根公式:

若

a = 29, b = 40, c = 12

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 29 \cdot 12}{2 \cdot 29}

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 29 \cdot 12}{2 \cdot 29}

4 0^{2} - 4 \cdot 29 \cdot 12 = 413

4 0^{2} - 4 \cdot 29 \cdot 12

数字相乘:

4 \cdot 29 \cdot 12 = 1392

= 4 0^{2} - 1392

4 0^{2} = 1600

= 1600 - 1392

数字相减:

1600 - 1392 = 208

= 208

208

质因数分解:

2^{4} \cdot 13

208

208

除以

2208 = 104 \cdot 2

= 2 \cdot 104

104

除以

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 52

52

除以

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 26

26

除以

226 = 13 \cdot 2

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

2, 13

都是质数，因此无法进一步因数分解

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

= 2^{4} \cdot 13

= 2^{4} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{4}

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} 13

整理后得

= 413

u_{1, 2} = \frac{- 40 \pm 4 13}{2 \cdot 29}

将解分隔开

u_{1} = \frac{- 40 + 4 13}{2 \cdot 29}, u_{2} = \frac{- 40 - 4 13}{2 \cdot 29}

u = \frac{- 40 + 4 13}{2 \cdot 29} : \frac{2 ( 13 - 10 )}{29}

\frac{- 40 + 4 13}{2 \cdot 29}

数字相乘:

2 \cdot 29 = 58

= \frac{- 40 + 4 13}{58}

分解

- 40 + 413 : 4 (- 10 + 13)

- 40 + 413

改写为

= - 4 \cdot 10 + 413

因式分解出通项

4

= 4 (- 10 + 13)

= \frac{4 ( - 10 + 13 )}{58}

约分:

2

= \frac{2 ( 13 - 10 )}{29}

u = \frac{- 40 - 4 13}{2 \cdot 29} : - \frac{2 ( 10 + 13 )}{29}

\frac{- 40 - 4 13}{2 \cdot 29}

数字相乘:

2 \cdot 29 = 58

= \frac{- 40 - 4 13}{58}

分解

- 40 - 413 : - 4 (10 + 13)

- 40 - 413

改写为

= - 4 \cdot 10 - 413

因式分解出通项

4

= - 4 (10 + 13)

= - \frac{4 ( 10 + 13 )}{58}

约分:

2

= - \frac{2 ( 10 + 13 )}{29}

二次方程组的解是:

u = \frac{2 ( 13 - 10 )}{29}, u = - \frac{2 ( 10 + 13 )}{29}

u = cos (x)

代回

cos (x) = \frac{2 ( 13 - 10 )}{29}, cos (x) = - \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 13 - 10 )}{29}, cos (x) = - \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 13 - 10 )}{29} : x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

cos (x) = \frac{2 ( 13 - 10 )}{29}

使用反三角函数性质

cos (x) = \frac{2 ( 13 - 10 )}{29}

cos (x) = \frac{2 ( 13 - 10 )}{29}

的通解

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

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

cos (x) = - \frac{2 ( 10 + 13 )}{29} : x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

cos (x) = - \frac{2 ( 10 + 13 )}{29}

使用反三角函数性质

cos (x) = - \frac{2 ( 10 + 13 )}{29}

cos (x) = - \frac{2 ( 10 + 13 )}{29}

的通解

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

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

合并所有解

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

将解代入原方程进行验证

将它们代入

5 cos (x) + 4 - 2 sin (x) = 0

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

的解:

真

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

对于

5 cos (x) + 4 - 2 sin (x) = 0

代入

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

5 cos (arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) + 4 - 2 sin (arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) = 0

整理后得

0 = 0

\Rightarrow 真

检验

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

的解:

假

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

代入

n = 1

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

对于

5 cos (x) + 4 - 2 sin (x) = 0

代入

x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

5 cos (- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) + 4 - 2 sin (- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) = 0

整理后得

3.59003 \dots = 0

\Rightarrow 假

检验

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

的解:

假

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

对于

5 cos (x) + 4 - 2 sin (x) = 0

代入

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

5 cos (arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 4 - 2 sin (arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 0

整理后得

- 1.38313 \dots = 0

\Rightarrow 假

检验

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

的解:

真

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

对于

5 cos (x) + 4 - 2 sin (x) = 0

代入

x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

5 cos (- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 4 - 2 sin (- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 0

整理后得

0 = 0

\Rightarrow 真

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

以小数形式表示解

x = 2.02750 \dots + 2 πn, x = - 2.78851 \dots + 2 πn

作图

流行的例子

sin (x) = - 0.54

tan (2 θ) = 1.82

sin (x) = - 0.65

sin (x) = - 0.71

sin (x) = - 0.98