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受欢迎的三角函数 >

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

解答

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

解答

x = - 2.71756 \dots + 2 πn, x = 1.14676 \dots + 2 πn

+1

度数

x = - 155.70481 \dots^{\circ} + 36 0^{\circ} n, x = 65.70481 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

4 cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

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

化简

- 4 - 16 cos (x) - 16 cos^{2} (x) + 16 (1 - cos^{2} (x)) : - 32 cos^{2} (x) - 16 cos (x) + 12

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

化简

- 4 - 16 cos (x) - 16 cos^{2} (x) + 16 - 16 cos^{2} (x) : - 32 cos^{2} (x) - 16 cos (x) + 12

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

对同类项分组

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

同类项相加:

- 16 cos^{2} (x) - 16 cos^{2} (x) = - 32 cos^{2} (x)

= - 16 cos (x) - 32 cos^{2} (x) - 4 + 16

数字相加/相减:

- 4 + 16 = 12

= - 32 cos^{2} (x) - 16 cos (x) + 12

= - 32 cos^{2} (x) - 16 cos (x) + 12

= - 32 cos^{2} (x) - 16 cos (x) + 12

12 - 16 cos (x) - 32 cos^{2} (x) = 0

用替代法求解

12 - 16 cos (x) - 32 cos^{2} (x) = 0

令

: cos (x) = u

12 - 16 u - 32 u^{2} = 0

12 - 16 u - 32 u^{2} = 0 : u = - \frac{1 + 7}{4}, u = \frac{7 - 1}{4}

12 - 16 u - 32 u^{2} = 0

改写成标准形式

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

- 32 u^{2} - 16 u + 12 = 0

使用求根公式求解

- 32 u^{2} - 16 u + 12 = 0

二次方程求根公式:

若

a = - 32, b = - 16, c = 12

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 32 ) \cdot 12}{2 ( - 32 )}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 32 ) \cdot 12}{2 ( - 32 )}

(- 16)^{2} - 4 (- 32) \cdot 12 = 167

(- 16)^{2} - 4 (- 32) \cdot 12

使用法则

- (- a) = a

= (- 16)^{2} + 4 \cdot 32 \cdot 12

使用指数法则:

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

若

n

是偶数

(- 16)^{2} = 1 6^{2}

= 1 6^{2} + 4 \cdot 32 \cdot 12

数字相乘:

4 \cdot 32 \cdot 12 = 1536

= 1 6^{2} + 1536

1 6^{2} = 256

= 256 + 1536

数字相加:

256 + 1536 = 1792

= 1792

1792

质因数分解:

2^{8} \cdot 7

1792

1792

除以

21792 = 896 \cdot 2

= 2 \cdot 896

896

除以

2896 = 448 \cdot 2

= 2 \cdot 2 \cdot 448

448

除以

2448 = 224 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 224

224

除以

2224 = 112 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 112

112

除以

2112 = 56 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 56

56

除以

256 = 28 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 28

28

除以

228 = 14 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7

2, 7

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 2 \cdot 2 \cdot 7

= 2^{8} \cdot 7

= 2^{8} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{8}

使用根式运算法则:

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

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 7

整理后得

= 167

u_{1, 2} = \frac{- ( - 16 ) \pm 16 7}{2 ( - 32 )}

将解分隔开

u_{1} = \frac{- ( - 16 ) + 16 7}{2 ( - 32 )}, u_{2} = \frac{- ( - 16 ) - 16 7}{2 ( - 32 )}

u = \frac{- ( - 16 ) + 16 7}{2 ( - 32 )} : - \frac{1 + 7}{4}

\frac{- ( - 16 ) + 16 7}{2 ( - 32 )}

去除括号:

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

= \frac{16 + 16 7}{- 2 \cdot 32}

数字相乘:

2 \cdot 32 = 64

= \frac{16 + 16 7}{- 64}

使用分式法则:

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

= - \frac{16 + 16 7}{64}

消掉

\frac{16 + 16 7}{64} : \frac{1 + 7}{4}

\frac{16 + 16 7}{64}

分解

16 + 167 : 16 (1 + 7)

16 + 167

改写为

= 16 \cdot 1 + 167

因式分解出通项

16

= 16 (1 + 7)

= \frac{16 ( 1 + 7 )}{64}

约分:

16

= \frac{1 + 7}{4}

= - \frac{1 + 7}{4}

u = \frac{- ( - 16 ) - 16 7}{2 ( - 32 )} : \frac{7 - 1}{4}

\frac{- ( - 16 ) - 16 7}{2 ( - 32 )}

去除括号:

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

= \frac{16 - 16 7}{- 2 \cdot 32}

数字相乘:

2 \cdot 32 = 64

= \frac{16 - 16 7}{- 64}

使用分式法则:

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

16 - 167 = - (167 - 16)

= \frac{16 7 - 16}{64}

分解

167 - 16 : 16 (7 - 1)

167 - 16

改写为

= 167 - 16 \cdot 1

因式分解出通项

16

= 16 (7 - 1)

= \frac{16 ( 7 - 1 )}{64}

约分:

16

= \frac{7 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 7}{4}, u = \frac{7 - 1}{4}

u = cos (x)

代回

cos (x) = - \frac{1 + 7}{4}, cos (x) = \frac{7 - 1}{4}

cos (x) = - \frac{1 + 7}{4}, cos (x) = \frac{7 - 1}{4}

cos (x) = - \frac{1 + 7}{4} : x = arccos (- \frac{1 + 7}{4}) + 2 πn, x = - arccos (- \frac{1 + 7}{4}) + 2 πn

cos (x) = - \frac{1 + 7}{4}

使用反三角函数性质

cos (x) = - \frac{1 + 7}{4}

cos (x) = - \frac{1 + 7}{4}

的通解

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

x = arccos (- \frac{1 + 7}{4}) + 2 πn, x = - arccos (- \frac{1 + 7}{4}) + 2 πn

x = arccos (- \frac{1 + 7}{4}) + 2 πn, x = - arccos (- \frac{1 + 7}{4}) + 2 πn

cos (x) = \frac{7 - 1}{4} : x = arccos (\frac{7 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

cos (x) = \frac{7 - 1}{4}

使用反三角函数性质

cos (x) = \frac{7 - 1}{4}

cos (x) = \frac{7 - 1}{4}

的通解

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

x = arccos (\frac{7 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

x = arccos (\frac{7 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

合并所有解

x = arccos (- \frac{1 + 7}{4}) + 2 πn, x = - arccos (- \frac{1 + 7}{4}) + 2 πn, x = arccos (\frac{7 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{1 + 7}{4}) + 2 πn

的解:

假

arccos (- \frac{1 + 7}{4}) + 2 πn

代入

n = 1

arccos (- \frac{1 + 7}{4}) + 2 π 1

对于

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

代入

x = arccos (- \frac{1 + 7}{4}) + 2 π 1

4 sin (arccos (- \frac{1 + 7}{4}) + 2 π 1) - 4 cos (arccos (- \frac{1 + 7}{4}) + 2 π 1) = 2

整理后得

5.29150 \dots = 2

\Rightarrow 假

检验

- arccos (- \frac{1 + 7}{4}) + 2 πn

的解:

真

- arccos (- \frac{1 + 7}{4}) + 2 πn

代入

n = 1

- arccos (- \frac{1 + 7}{4}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{1 + 7}{4}) + 2 π 1

4 sin (- arccos (- \frac{1 + 7}{4}) + 2 π 1) - 4 cos (- arccos (- \frac{1 + 7}{4}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

arccos (\frac{7 - 1}{4}) + 2 πn

的解:

真

arccos (\frac{7 - 1}{4}) + 2 πn

代入

n = 1

arccos (\frac{7 - 1}{4}) + 2 π 1

对于

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

代入

x = arccos (\frac{7 - 1}{4}) + 2 π 1

4 sin (arccos (\frac{7 - 1}{4}) + 2 π 1) - 4 cos (arccos (\frac{7 - 1}{4}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

2 π - arccos (\frac{7 - 1}{4}) + 2 πn

的解:

假

2 π - arccos (\frac{7 - 1}{4}) + 2 πn

代入

n = 1

2 π - arccos (\frac{7 - 1}{4}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{7 - 1}{4}) + 2 π 1

4 sin (2 π - arccos (\frac{7 - 1}{4}) + 2 π 1) - 4 cos (2 π - arccos (\frac{7 - 1}{4}) + 2 π 1) = 2

整理后得

- 5.29150 \dots = 2

\Rightarrow 假

x = - arccos (- \frac{1 + 7}{4}) + 2 πn, x = arccos (\frac{7 - 1}{4}) + 2 πn

以小数形式表示解

x = - 2.71756 \dots + 2 πn, x = 1.14676 \dots + 2 πn

作图

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