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受欢迎的三角函数 >

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

解答

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

解答

x = 1.11408 \dots + 2 πn, x = 2 π - 0.35307 \dots + 2 πn

+1

度数

x = 63.83252 \dots^{\circ} + 36 0^{\circ} n, x = 339.77029 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

5 cos (x)

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

两边进行平方

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

两边减去

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

4 sin^{2} (x) - 16 + 40 cos (x) - 25 cos^{2} (x) = 0

使用三角恒等式改写

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

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

化简

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 (1 - cos^{2} (x)) : 40 cos (x) - 29 cos^{2} (x) - 12

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

- 16 + 4 = - 12

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

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

= 40 cos (x) - 29 cos^{2} (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 ( 10 - 13 )}{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 ( - 29 ) ( - 12 )}{2 ( - 29 )}

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

4 0^{2} - 4 (- 29) (- 12) = 413

4 0^{2} - 4 (- 29) (- 12)

使用法则

- (- a) = a

= 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 ( - 29 )}

将解分隔开

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

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

\frac{- 40 + 4 13}{2 ( - 29 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 29 = 58

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

使用分式法则:

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

- 40 + 413 = - (40 - 413)

= \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}

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

\frac{- 40 - 4 13}{2 ( - 29 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 29 = 58

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

使用分式法则:

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

- 40 - 413 = - (40 + 413)

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

u = cos (x)

代回

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

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

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

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

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

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

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

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

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

代入

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

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

整理后得

4 = 4

\Rightarrow 真

检验

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

的解:

假

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

代入

n = 1

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

对于

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

代入

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

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

整理后得

0.40996 \dots = 4

\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

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

代入

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

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

整理后得

5.38313 \dots = 4

\Rightarrow 假

检验

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

的解:

真

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

代入

n = 1

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

对于

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

代入

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

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

整理后得

4 = 4

\Rightarrow 真

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

以小数形式表示解

x = 1.11408 \dots + 2 πn, x = 2 π - 0.35307 \dots + 2 πn

作图

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