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受欢迎的三角函数 >

50sin(x)+15cos(x)=40,0<x<pi

解答

50 sin (x) + 15 cos (x) = 40, 0 < x < π

解答

x = 1.97713 \dots, x = 0.58154 \dots

+1

度数

x = 113.28144 \dots^{\circ}, x = 33.32006 \dots^{\circ}

求解步骤

50 sin (x) + 15 cos (x) = 40, 0 < x < π

两边减去

15 cos (x)

50 sin (x) = 40 - 15 cos (x)

两边进行平方

(50 sin (x))^{2} = (40 - 15 cos (x))^{2}

两边减去

(40 - 15 cos (x))^{2}

2500 sin^{2} (x) - 1600 + 1200 cos (x) - 225 cos^{2} (x) = 0

使用三角恒等式改写

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 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)

= - 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x))

化简

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x)) : 1200 cos (x) - 2725 cos^{2} (x) + 900

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 (1 - cos^{2} (x))

乘开

2500 (1 - cos^{2} (x)) : 2500 - 2500 cos^{2} (x)

2500 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2500, b = 1, c = cos^{2} (x)

= 2500 \cdot 1 - 2500 cos^{2} (x)

数字相乘:

2500 \cdot 1 = 2500

= 2500 - 2500 cos^{2} (x)

= - 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

化简

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x) : 1200 cos (x) - 2725 cos^{2} (x) + 900

- 1600 + 1200 cos (x) - 225 cos^{2} (x) + 2500 - 2500 cos^{2} (x)

对同类项分组

= 1200 cos (x) - 225 cos^{2} (x) - 2500 cos^{2} (x) - 1600 + 2500

同类项相加:

- 225 cos^{2} (x) - 2500 cos^{2} (x) = - 2725 cos^{2} (x)

= 1200 cos (x) - 2725 cos^{2} (x) - 1600 + 2500

数字相加/相减:

- 1600 + 2500 = 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

= 1200 cos (x) - 2725 cos^{2} (x) + 900

900 + 1200 cos (x) - 2725 cos^{2} (x) = 0

用替代法求解

900 + 1200 cos (x) - 2725 cos^{2} (x) = 0

令

: cos (x) = u

900 + 1200 u - 2725 u^{2} = 0

900 + 1200 u - 2725 u^{2} = 0 : u = - \frac{6 ( 5 5 - 4 )}{109}, u = \frac{6 ( 4 + 5 5 )}{109}

900 + 1200 u - 2725 u^{2} = 0

改写成标准形式

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

- 2725 u^{2} + 1200 u + 900 = 0

使用求根公式求解

- 2725 u^{2} + 1200 u + 900 = 0

二次方程求根公式:

若

a = - 2725, b = 1200, c = 900

u_{1, 2} = \frac{- 1200 \pm 120 0 ^{2} - 4 ( - 2725 ) \cdot 900}{2 ( - 2725 )}

u_{1, 2} = \frac{- 1200 \pm 120 0 ^{2} - 4 ( - 2725 ) \cdot 900}{2 ( - 2725 )}

120 0^{2} - 4 (- 2725) \cdot 900 = 15005

120 0^{2} - 4 (- 2725) \cdot 900

使用法则

- (- a) = a

= 120 0^{2} + 4 \cdot 2725 \cdot 900

数字相乘:

4 \cdot 2725 \cdot 900 = 9810000

= 120 0^{2} + 9810000

120 0^{2} = 1440000

= 1440000 + 9810000

数字相加:

1440000 + 9810000 = 11250000

= 11250000

11250000

质因数分解:

2^{4} \cdot 3^{2} \cdot 5^{7}

11250000

= 5^{7} \cdot 2^{4} \cdot 3^{2}

使用指数法则:

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

= 5^{6} \cdot 2^{4} \cdot 3^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{4} 3^{2} 5^{6}

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} 5 3^{2} 5^{6}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{2} \cdot 35 5^{6}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

5^{6} = 5^{\frac{6}{2}} = 5^{3}

= 5^{3} \cdot 2^{2} \cdot 35

整理后得

= 15005

u_{1, 2} = \frac{- 1200 \pm 1500 5}{2 ( - 2725 )}

将解分隔开

u_{1} = \frac{- 1200 + 1500 5}{2 ( - 2725 )}, u_{2} = \frac{- 1200 - 1500 5}{2 ( - 2725 )}

u = \frac{- 1200 + 1500 5}{2 ( - 2725 )} : - \frac{6 ( 5 5 - 4 )}{109}

\frac{- 1200 + 1500 5}{2 ( - 2725 )}

去除括号:

(- a) = - a

= \frac{- 1200 + 1500 5}{- 2 \cdot 2725}

数字相乘:

2 \cdot 2725 = 5450

= \frac{- 1200 + 1500 5}{- 5450}

使用分式法则:

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

= - \frac{- 1200 + 1500 5}{5450}

消掉

\frac{- 1200 + 1500 5}{5450} : \frac{6 ( 5 5 - 4 )}{109}

\frac{- 1200 + 1500 5}{5450}

分解

- 1200 + 15005 : 300 (- 4 + 55)

- 1200 + 15005

改写为

= - 300 \cdot 4 + 300 \cdot 55

因式分解出通项

300

= 300 (- 4 + 55)

= \frac{300 ( - 4 + 5 5 )}{5450}

约分:

50

= \frac{6 ( 5 5 - 4 )}{109}

= - \frac{6 ( 5 5 - 4 )}{109}

u = \frac{- 1200 - 1500 5}{2 ( - 2725 )} : \frac{6 ( 4 + 5 5 )}{109}

\frac{- 1200 - 1500 5}{2 ( - 2725 )}

去除括号:

(- a) = - a

= \frac{- 1200 - 1500 5}{- 2 \cdot 2725}

数字相乘:

2 \cdot 2725 = 5450

= \frac{- 1200 - 1500 5}{- 5450}

使用分式法则:

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

- 1200 - 15005 = - (1200 + 15005)

= \frac{1200 + 1500 5}{5450}

分解

1200 + 15005 : 300 (4 + 55)

1200 + 15005

改写为

= 300 \cdot 4 + 300 \cdot 55

因式分解出通项

300

= 300 (4 + 55)

= \frac{300 ( 4 + 5 5 )}{5450}

约分:

50

= \frac{6 ( 4 + 5 5 )}{109}

二次方程组的解是:

u = - \frac{6 ( 5 5 - 4 )}{109}, u = \frac{6 ( 4 + 5 5 )}{109}

u = cos (x)

代回

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, cos (x) = \frac{6 ( 4 + 5 5 )}{109}

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, cos (x) = \frac{6 ( 4 + 5 5 )}{109}

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, 0 < x < π : x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}, 0 < x < π

使用反三角函数性质

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}

cos (x) = - \frac{6 ( 5 5 - 4 )}{109}

的通解

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

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn, x = - arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn, x = - arccos (- \frac{6 ( 5 5 - 4 )}{109}) + 2 πn

在

0 < x < π

范围内的解

x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

cos (x) = \frac{6 ( 4 + 5 5 )}{109}, 0 < x < π : x = arccos (\frac{6 ( 4 + 5 5 )}{109})

cos (x) = \frac{6 ( 4 + 5 5 )}{109}, 0 < x < π

使用反三角函数性质

cos (x) = \frac{6 ( 4 + 5 5 )}{109}

cos (x) = \frac{6 ( 4 + 5 5 )}{109}

的通解

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

x = arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn, x = 2 π - arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn

x = arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn, x = 2 π - arccos (\frac{6 ( 4 + 5 5 )}{109}) + 2 πn

在

0 < x < π

范围内的解

x = arccos (\frac{6 ( 4 + 5 5 )}{109})

合并所有解

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}), x = arccos (\frac{6 ( 4 + 5 5 )}{109})

将解代入原方程进行验证

将它们代入

50 sin (x) + 15 cos (x) = 40

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{6 ( 5 5 - 4 )}{109})

的解:

真

arccos (- \frac{6 ( 5 5 - 4 )}{109})

代入

n = 1

arccos (- \frac{6 ( 5 5 - 4 )}{109})

对于

50 sin (x) + 15 cos (x) = 40

代入

x = arccos (- \frac{6 ( 5 5 - 4 )}{109})

50 sin (arccos (- \frac{6 ( 5 5 - 4 )}{109})) + 15 cos (arccos (- \frac{6 ( 5 5 - 4 )}{109})) = 40

整理后得

40 = 40

\Rightarrow 真

检验

arccos (\frac{6 ( 4 + 5 5 )}{109})

的解:

真

arccos (\frac{6 ( 4 + 5 5 )}{109})

代入

n = 1

arccos (\frac{6 ( 4 + 5 5 )}{109})

对于

50 sin (x) + 15 cos (x) = 40

代入

x = arccos (\frac{6 ( 4 + 5 5 )}{109})

50 sin (arccos (\frac{6 ( 4 + 5 5 )}{109})) + 15 cos (arccos (\frac{6 ( 4 + 5 5 )}{109})) = 40

整理后得

40 = 40

\Rightarrow 真

x = arccos (- \frac{6 ( 5 5 - 4 )}{109}), x = arccos (\frac{6 ( 4 + 5 5 )}{109})

以小数形式表示解

x = 1.97713 \dots, x = 0.58154 \dots

作图

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