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受欢迎的三角函数 >

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

解答

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

解答

x = 2.88012 \dots + 2 πn, x = 0.75142 \dots + 2 πn

+1

度数

x = 165.01910 \dots^{\circ} + 36 0^{\circ} n, x = 43.05338 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

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

化简

- 4 - cos^{2} (x) + 16 (1 - cos^{2} (x)) - 4 cos (x) : - 17 cos^{2} (x) - 4 cos (x) + 12

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

乘开

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

数字相乘:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

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

化简

- 4 - cos^{2} (x) + 16 - 16 cos^{2} (x) - 4 cos (x) : - 17 cos^{2} (x) - 4 cos (x) + 12

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

对同类项分组

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

同类项相加:

- cos^{2} (x) - 16 cos^{2} (x) = - 17 cos^{2} (x)

= - 17 cos^{2} (x) - 4 cos (x) - 4 + 16

数字相加/相减:

- 4 + 16 = 12

= - 17 cos^{2} (x) - 4 cos (x) + 12

= - 17 cos^{2} (x) - 4 cos (x) + 12

= - 17 cos^{2} (x) - 4 cos (x) + 12

12 - 17 cos^{2} (x) - 4 cos (x) = 0

用替代法求解

12 - 17 cos^{2} (x) - 4 cos (x) = 0

令

: cos (x) = u

12 - 17 u^{2} - 4 u = 0

12 - 17 u^{2} - 4 u = 0 : u = - \frac{2 ( 1 + 2 13 )}{17}, u = \frac{2 ( 2 13 - 1 )}{17}

12 - 17 u^{2} - 4 u = 0

改写成标准形式

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

- 17 u^{2} - 4 u + 12 = 0

使用求根公式求解

- 17 u^{2} - 4 u + 12 = 0

二次方程求根公式:

若

a = - 17, b = - 4, c = 12

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 17 ) \cdot 12}{2 ( - 17 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 17 ) \cdot 12}{2 ( - 17 )}

(- 4)^{2} - 4 (- 17) \cdot 12 = 813

(- 4)^{2} - 4 (- 17) \cdot 12

使用法则

- (- a) = a

= (- 4)^{2} + 4 \cdot 17 \cdot 12

使用指数法则:

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

若

n

是偶数

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

= 4^{2} + 4 \cdot 17 \cdot 12

数字相乘:

4 \cdot 17 \cdot 12 = 816

= 4^{2} + 816

4^{2} = 16

= 16 + 816

数字相加:

16 + 816 = 832

= 832

832

质因数分解:

2^{6} \cdot 13

832

832

除以

2832 = 416 \cdot 2

= 2 \cdot 416

416

除以

2416 = 208 \cdot 2

= 2 \cdot 2 \cdot 208

208

除以

2208 = 104 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 104

104

除以

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 52

52

除以

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 26

26

除以

226 = 13 \cdot 2

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

= 2^{6} \cdot 13

= 2^{6} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{6}

使用根式运算法则:

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

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

= 2^{3} 13

整理后得

= 813

u_{1, 2} = \frac{- ( - 4 ) \pm 8 13}{2 ( - 17 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 8 13}{2 ( - 17 )}, u_{2} = \frac{- ( - 4 ) - 8 13}{2 ( - 17 )}

u = \frac{- ( - 4 ) + 8 13}{2 ( - 17 )} : - \frac{2 ( 1 + 2 13 )}{17}

\frac{- ( - 4 ) + 8 13}{2 ( - 17 )}

去除括号:

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

= \frac{4 + 8 13}{- 2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{4 + 8 13}{- 34}

使用分式法则:

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

= - \frac{4 + 8 13}{34}

消掉

\frac{4 + 8 13}{34} : \frac{2 ( 1 + 2 13 )}{17}

\frac{4 + 8 13}{34}

分解

4 + 813 : 4 (1 + 213)

4 + 813

改写为

= 4 \cdot 1 + 4 \cdot 213

因式分解出通项

4

= 4 (1 + 213)

= \frac{4 ( 1 + 2 13 )}{34}

约分:

2

= \frac{2 ( 1 + 2 13 )}{17}

= - \frac{2 ( 1 + 2 13 )}{17}

u = \frac{- ( - 4 ) - 8 13}{2 ( - 17 )} : \frac{2 ( 2 13 - 1 )}{17}

\frac{- ( - 4 ) - 8 13}{2 ( - 17 )}

去除括号:

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

= \frac{4 - 8 13}{- 2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{4 - 8 13}{- 34}

使用分式法则:

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

4 - 813 = - (813 - 4)

= \frac{8 13 - 4}{34}

分解

813 - 4 : 4 (213 - 1)

813 - 4

改写为

= 4 \cdot 213 - 4 \cdot 1

因式分解出通项

4

= 4 (213 - 1)

= \frac{4 ( 2 13 - 1 )}{34}

约分:

2

= \frac{2 ( 2 13 - 1 )}{17}

二次方程组的解是:

u = - \frac{2 ( 1 + 2 13 )}{17}, u = \frac{2 ( 2 13 - 1 )}{17}

u = cos (x)

代回

cos (x) = - \frac{2 ( 1 + 2 13 )}{17}, cos (x) = \frac{2 ( 2 13 - 1 )}{17}

cos (x) = - \frac{2 ( 1 + 2 13 )}{17}, cos (x) = \frac{2 ( 2 13 - 1 )}{17}

cos (x) = - \frac{2 ( 1 + 2 13 )}{17} : x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

cos (x) = - \frac{2 ( 1 + 2 13 )}{17}

使用反三角函数性质

cos (x) = - \frac{2 ( 1 + 2 13 )}{17}

cos (x) = - \frac{2 ( 1 + 2 13 )}{17}

的通解

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

x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

cos (x) = \frac{2 ( 2 13 - 1 )}{17} : x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

cos (x) = \frac{2 ( 2 13 - 1 )}{17}

使用反三角函数性质

cos (x) = \frac{2 ( 2 13 - 1 )}{17}

cos (x) = \frac{2 ( 2 13 - 1 )}{17}

的通解

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

x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

合并所有解

x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, x = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

的解:

真

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

对于

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

代入

x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

4 sin (arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) = cos (arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) + 2

整理后得

1.03398 \dots = 1.03398 \dots

\Rightarrow 真

检验

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

的解:

假

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

4 sin (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) = cos (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) + 2

整理后得

- 1.03398 \dots = 1.03398 \dots

\Rightarrow 假

检验

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

的解:

真

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

对于

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

代入

x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

4 sin (arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) = cos (arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) + 2

整理后得

2.73071 \dots = 2.73071 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

4 sin (2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) = cos (2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) + 2

整理后得

- 2.73071 \dots = 2.73071 \dots

\Rightarrow 假

x = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, x = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

以小数形式表示解

x = 2.88012 \dots + 2 πn, x = 0.75142 \dots + 2 πn

作图

流行的例子

tan(β+10)=cot(2β-10)

tan (β + 1 0^{\circ}) = cot (2 β - 1 0^{\circ})

1-2cos^2(8x)=sin(4x)

1 - 2 cos^{2} (8 x) = sin (4 x)

tan (x - 1 0^{\circ}) = 0

cos^2(x)=3sin(x)cos(x)

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

sin(y)=(50)/(65.3)

sin (y) = \frac{50}{65.3}