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受欢迎的三角函数 >

5sin(x)+12cos(x)=13

解答

5 sin (x) + 12 cos (x) = 13

解答

x = 0.39479 \dots + 2 πn

+1

度数

x = 22.61986 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 sin (x) + 12 cos (x) = 13

两边减去

12 cos (x)

5 sin (x) = 13 - 12 cos (x)

两边进行平方

(5 sin (x))^{2} = (13 - 12 cos (x))^{2}

两边减去

(13 - 12 cos (x))^{2}

25 sin^{2} (x) - 169 + 312 cos (x) - 144 cos^{2} (x) = 0

使用三角恒等式改写

- 169 - 144 cos^{2} (x) + 25 sin^{2} (x) + 312 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)

= - 169 - 144 cos^{2} (x) + 25 (1 - cos^{2} (x)) + 312 cos (x)

化简

- 169 - 144 cos^{2} (x) + 25 (1 - cos^{2} (x)) + 312 cos (x) : 312 cos (x) - 169 cos^{2} (x) - 144

- 169 - 144 cos^{2} (x) + 25 (1 - cos^{2} (x)) + 312 cos (x)

乘开

25 (1 - cos^{2} (x)) : 25 - 25 cos^{2} (x)

25 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 25, b = 1, c = cos^{2} (x)

= 25 \cdot 1 - 25 cos^{2} (x)

数字相乘:

25 \cdot 1 = 25

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

= - 169 - 144 cos^{2} (x) + 25 - 25 cos^{2} (x) + 312 cos (x)

化简

- 169 - 144 cos^{2} (x) + 25 - 25 cos^{2} (x) + 312 cos (x) : 312 cos (x) - 169 cos^{2} (x) - 144

- 169 - 144 cos^{2} (x) + 25 - 25 cos^{2} (x) + 312 cos (x)

对同类项分组

= - 144 cos^{2} (x) - 25 cos^{2} (x) + 312 cos (x) - 169 + 25

同类项相加:

- 144 cos^{2} (x) - 25 cos^{2} (x) = - 169 cos^{2} (x)

= - 169 cos^{2} (x) + 312 cos (x) - 169 + 25

数字相加/相减:

- 169 + 25 = - 144

= 312 cos (x) - 169 cos^{2} (x) - 144

= 312 cos (x) - 169 cos^{2} (x) - 144

= 312 cos (x) - 169 cos^{2} (x) - 144

- 144 - 169 cos^{2} (x) + 312 cos (x) = 0

用替代法求解

- 144 - 169 cos^{2} (x) + 312 cos (x) = 0

令

: cos (x) = u

- 144 - 169 u^{2} + 312 u = 0

- 144 - 169 u^{2} + 312 u = 0 : u = \frac{12}{13}

- 144 - 169 u^{2} + 312 u = 0

改写成标准形式

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

- 169 u^{2} + 312 u - 144 = 0

使用求根公式求解

- 169 u^{2} + 312 u - 144 = 0

二次方程求根公式:

若

a = - 169, b = 312, c = - 144

u_{1, 2} = \frac{- 312 \pm 31 2 ^{2} - 4 ( - 169 ) ( - 144 )}{2 ( - 169 )}

u_{1, 2} = \frac{- 312 \pm 31 2 ^{2} - 4 ( - 169 ) ( - 144 )}{2 ( - 169 )}

31 2^{2} - 4 (- 169) (- 144) = 0

31 2^{2} - 4 (- 169) (- 144)

使用法则

- (- a) = a

= 31 2^{2} - 4 \cdot 169 \cdot 144

数字相乘:

4 \cdot 169 \cdot 144 = 97344

= 31 2^{2} - 97344

31 2^{2} = 97344

= 97344 - 97344

数字相减:

97344 - 97344 = 0

= 0

u_{1, 2} = \frac{- 312 \pm 0}{2 ( - 169 )}

u = \frac{- 312}{2 ( - 169 )}

\frac{- 312}{2 ( - 169 )} = \frac{12}{13}

\frac{- 312}{2 ( - 169 )}

去除括号:

(- a) = - a

= \frac{- 312}{- 2 \cdot 169}

数字相乘:

2 \cdot 169 = 338

= \frac{- 312}{- 338}

使用分式法则:

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

= \frac{312}{338}

约分:

26

= \frac{12}{13}

u = \frac{12}{13}

二次方程组的解是:

u = \frac{12}{13}

u = cos (x)

代回

cos (x) = \frac{12}{13}

cos (x) = \frac{12}{13}

cos (x) = \frac{12}{13} : x = arccos (\frac{12}{13}) + 2 πn, x = 2 π - arccos (\frac{12}{13}) + 2 πn

cos (x) = \frac{12}{13}

使用反三角函数性质

cos (x) = \frac{12}{13}

cos (x) = \frac{12}{13}

的通解

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

x = arccos (\frac{12}{13}) + 2 πn, x = 2 π - arccos (\frac{12}{13}) + 2 πn

x = arccos (\frac{12}{13}) + 2 πn, x = 2 π - arccos (\frac{12}{13}) + 2 πn

合并所有解

x = arccos (\frac{12}{13}) + 2 πn, x = 2 π - arccos (\frac{12}{13}) + 2 πn

将解代入原方程进行验证

将它们代入

5 sin (x) + 12 cos (x) = 13

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{12}{13}) + 2 πn

的解:

真

arccos (\frac{12}{13}) + 2 πn

代入

n = 1

arccos (\frac{12}{13}) + 2 π 1

对于

5 sin (x) + 12 cos (x) = 13

代入

x = arccos (\frac{12}{13}) + 2 π 1

5 sin (arccos (\frac{12}{13}) + 2 π 1) + 12 cos (arccos (\frac{12}{13}) + 2 π 1) = 13

整理后得

13 = 13

\Rightarrow 真

检验

2 π - arccos (\frac{12}{13}) + 2 πn

的解:

假

2 π - arccos (\frac{12}{13}) + 2 πn

代入

n = 1

2 π - arccos (\frac{12}{13}) + 2 π 1

对于

5 sin (x) + 12 cos (x) = 13

代入

x = 2 π - arccos (\frac{12}{13}) + 2 π 1

5 sin (2 π - arccos (\frac{12}{13}) + 2 π 1) + 12 cos (2 π - arccos (\frac{12}{13}) + 2 π 1) = 13

整理后得

9.15384 \dots = 13

\Rightarrow 假

x = arccos (\frac{12}{13}) + 2 πn

以小数形式表示解

x = 0.39479 \dots + 2 πn

作图

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