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受欢迎的三角函数 >

3sin(x)-2cos(x)=1

解答

3 sin (x) - 2 cos (x) = 1

解答

x = - 2.83462 \dots + 2 πn, x = 0.86903 \dots + 2 πn

+1

度数

x = - 162.41204 \dots^{\circ} + 36 0^{\circ} n, x = 49.79218 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (x) - 2 cos (x) = 1

两边加上

2 cos (x)

3 sin (x) = 1 + 2 cos (x)

两边进行平方

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

两边减去

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

9 sin^{2} (x) - 1 - 4 cos (x) - 4 cos^{2} (x) = 0

使用三角恒等式改写

- 1 - 4 cos (x) - 4 cos^{2} (x) + 9 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)

= - 1 - 4 cos (x) - 4 cos^{2} (x) + 9 (1 - cos^{2} (x))

化简

- 1 - 4 cos (x) - 4 cos^{2} (x) + 9 (1 - cos^{2} (x)) : - 13 cos^{2} (x) - 4 cos (x) + 8

- 1 - 4 cos (x) - 4 cos^{2} (x) + 9 (1 - cos^{2} (x))

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= - 1 - 4 cos (x) - 4 cos^{2} (x) + 9 - 9 cos^{2} (x)

化简

- 1 - 4 cos (x) - 4 cos^{2} (x) + 9 - 9 cos^{2} (x) : - 13 cos^{2} (x) - 4 cos (x) + 8

- 1 - 4 cos (x) - 4 cos^{2} (x) + 9 - 9 cos^{2} (x)

对同类项分组

= - 4 cos (x) - 4 cos^{2} (x) - 9 cos^{2} (x) - 1 + 9

同类项相加:

- 4 cos^{2} (x) - 9 cos^{2} (x) = - 13 cos^{2} (x)

= - 4 cos (x) - 13 cos^{2} (x) - 1 + 9

数字相加/相减:

- 1 + 9 = 8

= - 13 cos^{2} (x) - 4 cos (x) + 8

= - 13 cos^{2} (x) - 4 cos (x) + 8

= - 13 cos^{2} (x) - 4 cos (x) + 8

8 - 13 cos^{2} (x) - 4 cos (x) = 0

用替代法求解

8 - 13 cos^{2} (x) - 4 cos (x) = 0

令

: cos (x) = u

8 - 13 u^{2} - 4 u = 0

8 - 13 u^{2} - 4 u = 0 : u = - \frac{2 ( 1 + 3 3 )}{13}, u = \frac{2 ( 3 3 - 1 )}{13}

8 - 13 u^{2} - 4 u = 0

改写成标准形式

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

- 13 u^{2} - 4 u + 8 = 0

使用求根公式求解

- 13 u^{2} - 4 u + 8 = 0

二次方程求根公式:

若

a = - 13, b = - 4, c = 8

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 13 ) \cdot 8}{2 ( - 13 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 13 ) \cdot 8}{2 ( - 13 )}

(- 4)^{2} - 4 (- 13) \cdot 8 = 123

(- 4)^{2} - 4 (- 13) \cdot 8

使用法则

- (- a) = a

= (- 4)^{2} + 4 \cdot 13 \cdot 8

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 4^{2} + 4 \cdot 13 \cdot 8

数字相乘:

4 \cdot 13 \cdot 8 = 416

= 4^{2} + 416

4^{2} = 16

= 16 + 416

数字相加:

16 + 416 = 432

= 432

432

质因数分解:

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

432

432

除以

2432 = 216 \cdot 2

= 2 \cdot 216

216

除以

2216 = 108 \cdot 2

= 2 \cdot 2 \cdot 108

108

除以

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 54

54

除以

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 27

27

除以

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

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

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

使用指数法则:

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

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

使用根式运算法则:

n ab = n a n b

= 3 2^{4} 3^{2}

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} 3 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{2} \cdot 33

整理后得

= 123

u_{1, 2} = \frac{- ( - 4 ) \pm 12 3}{2 ( - 13 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 12 3}{2 ( - 13 )}, u_{2} = \frac{- ( - 4 ) - 12 3}{2 ( - 13 )}

u = \frac{- ( - 4 ) + 12 3}{2 ( - 13 )} : - \frac{2 ( 1 + 3 3 )}{13}

\frac{- ( - 4 ) + 12 3}{2 ( - 13 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{4 + 12 3}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{4 + 12 3}{- 26}

使用分式法则:

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

= - \frac{4 + 12 3}{26}

消掉

\frac{4 + 12 3}{26} : \frac{2 ( 1 + 3 3 )}{13}

\frac{4 + 12 3}{26}

分解

4 + 123 : 4 (1 + 33)

4 + 123

改写为

= 4 \cdot 1 + 4 \cdot 33

因式分解出通项

4

= 4 (1 + 33)

= \frac{4 ( 1 + 3 3 )}{26}

约分:

2

= \frac{2 ( 1 + 3 3 )}{13}

= - \frac{2 ( 1 + 3 3 )}{13}

u = \frac{- ( - 4 ) - 12 3}{2 ( - 13 )} : \frac{2 ( 3 3 - 1 )}{13}

\frac{- ( - 4 ) - 12 3}{2 ( - 13 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{4 - 12 3}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{4 - 12 3}{- 26}

使用分式法则:

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

4 - 123 = - (123 - 4)

= \frac{12 3 - 4}{26}

分解

123 - 4 : 4 (33 - 1)

123 - 4

改写为

= 4 \cdot 33 - 4 \cdot 1

因式分解出通项

4

= 4 (33 - 1)

= \frac{4 ( 3 3 - 1 )}{26}

约分:

2

= \frac{2 ( 3 3 - 1 )}{13}

二次方程组的解是:

u = - \frac{2 ( 1 + 3 3 )}{13}, u = \frac{2 ( 3 3 - 1 )}{13}

u = cos (x)

代回

cos (x) = - \frac{2 ( 1 + 3 3 )}{13}, cos (x) = \frac{2 ( 3 3 - 1 )}{13}

cos (x) = - \frac{2 ( 1 + 3 3 )}{13}, cos (x) = \frac{2 ( 3 3 - 1 )}{13}

cos (x) = - \frac{2 ( 1 + 3 3 )}{13} : x = arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

cos (x) = - \frac{2 ( 1 + 3 3 )}{13}

使用反三角函数性质

cos (x) = - \frac{2 ( 1 + 3 3 )}{13}

cos (x) = - \frac{2 ( 1 + 3 3 )}{13}

的通解

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

x = arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

x = arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

cos (x) = \frac{2 ( 3 3 - 1 )}{13} : x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

cos (x) = \frac{2 ( 3 3 - 1 )}{13}

使用反三角函数性质

cos (x) = \frac{2 ( 3 3 - 1 )}{13}

cos (x) = \frac{2 ( 3 3 - 1 )}{13}

的通解

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

x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

合并所有解

x = arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

将解代入原方程进行验证

将它们代入

3 sin (x) - 2 cos (x) = 1

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

的解:

假

arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 1

代入

x = arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

3 sin (arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) - 2 cos (arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) = 1

整理后得

2.81301 \dots = 1

\Rightarrow 假

检验

- arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

的解:

真

- arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 1

代入

x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

3 sin (- arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) - 2 cos (- arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

检验

arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

的解:

真

arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 1

代入

x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

3 sin (arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) - 2 cos (arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 1

代入

x = 2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

3 sin (2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) - 2 cos (2 π - arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) = 1

整理后得

- 3.58224 \dots = 1

\Rightarrow 假

x = - arccos (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arccos (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

以小数形式表示解

x = - 2.83462 \dots + 2 πn, x = 0.86903 \dots + 2 πn

作图

流行的例子

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40 cos (\frac{2 π}{3} t) - 71 = - 80

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