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受欢迎的三角函数 >

4cos(x)-4sin(x)=2sqrt(6)

解答

4 cos (x) - 4 sin (x) = 26

解答

x = - 1.30899 \dots + 2 πn, x = - 0.26179 \dots + 2 πn

+1

度数

x = - 7 5^{\circ} + 36 0^{\circ} n, x = - 1 5^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (x) - 4 sin (x) = 26

两边加上

4 sin (x)

4 cos (x) = 26 + 4 sin (x)

两边进行平方

(4 cos (x))^{2} = (26 + 4 sin (x))^{2}

两边减去

(26 + 4 sin (x))^{2}

16 cos^{2} (x) - 24 - 166 sin (x) - 16 sin^{2} (x) = 0

使用三角恒等式改写

- 24 + 16 cos^{2} (x) - 16 sin^{2} (x) - 16 sin (x) 6

使用毕达哥拉斯恒等式:

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

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

= - 24 + 16 (1 - sin^{2} (x)) - 16 sin^{2} (x) - 16 sin (x) 6

化简

- 24 + 16 (1 - sin^{2} (x)) - 16 sin^{2} (x) - 16 sin (x) 6 : - 32 sin^{2} (x) - 166 sin (x) - 8

- 24 + 16 (1 - sin^{2} (x)) - 16 sin^{2} (x) - 16 sin (x) 6

= - 24 + 16 (1 - sin^{2} (x)) - 16 sin^{2} (x) - 166 sin (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

16 \cdot 1 = 16

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

= - 24 + 16 - 16 sin^{2} (x) - 16 sin^{2} (x) - 16 sin (x) 6

化简

- 24 + 16 - 16 sin^{2} (x) - 16 sin^{2} (x) - 16 sin (x) 6 : - 32 sin^{2} (x) - 166 sin (x) - 8

- 24 + 16 - 16 sin^{2} (x) - 16 sin^{2} (x) - 16 sin (x) 6

同类项相加:

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

= - 24 + 16 - 32 sin^{2} (x) - 166 sin (x)

数字相加/相减:

- 24 + 16 = - 8

= - 32 sin^{2} (x) - 166 sin (x) - 8

= - 32 sin^{2} (x) - 166 sin (x) - 8

= - 32 sin^{2} (x) - 166 sin (x) - 8

- 8 - 32 sin^{2} (x) - 16 sin (x) 6 = 0

用替代法求解

- 8 - 32 sin^{2} (x) - 16 sin (x) 6 = 0

令

: sin (x) = u

- 8 - 32 u^{2} - 16 u 6 = 0

- 8 - 32 u^{2} - 16 u 6 = 0 : u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

- 8 - 32 u^{2} - 16 u 6 = 0

改写成标准形式

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

- 32 u^{2} - 166 u - 8 = 0

使用求根公式求解

- 32 u^{2} - 166 u - 8 = 0

二次方程求根公式:

若

a = - 32, b = - 166, c = - 8

u_{1, 2} = \frac{- ( - 16 6 ) \pm ( - 16 6 ) ^{2} - 4 ( - 32 ) ( - 8 )}{2 ( - 32 )}

u_{1, 2} = \frac{- ( - 16 6 ) \pm ( - 16 6 ) ^{2} - 4 ( - 32 ) ( - 8 )}{2 ( - 32 )}

(- 166)^{2} - 4 (- 32) (- 8) = 162

(- 166)^{2} - 4 (- 32) (- 8)

使用法则

- (- a) = a

= (- 166)^{2} - 4 \cdot 32 \cdot 8

(- 166)^{2} = 1 6^{2} \cdot 6

(- 166)^{2}

使用指数法则:

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

若

n

是偶数

(- 166)^{2} = (166)^{2}

= (166)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

(6)^{2} : 6

使用根式运算法则:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 6

= 1 6^{2} \cdot 6

4 \cdot 32 \cdot 8 = 1024

4 \cdot 32 \cdot 8

数字相乘:

4 \cdot 32 \cdot 8 = 1024

= 1024

= 1 6^{2} \cdot 6 - 1024

1 6^{2} \cdot 6 = 1536

1 6^{2} \cdot 6

1 6^{2} = 256

= 256 \cdot 6

数字相乘:

256 \cdot 6 = 1536

= 1536

= 1536 - 1024

数字相减:

1536 - 1024 = 512

= 512

512

质因数分解:

2^{9}

512

512

除以

2512 = 256 \cdot 2

= 2 \cdot 256

256

除以

2256 = 128 \cdot 2

= 2 \cdot 2 \cdot 128

128

除以

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 64

64

除以

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

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

2

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 2

= 2^{9}

= 2^{9}

使用指数法则:

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

= 2^{8} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 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} 2

整理后得

= 162

u_{1, 2} = \frac{- ( - 16 6 ) \pm 16 2}{2 ( - 32 )}

将解分隔开

u_{1} = \frac{- ( - 16 6 ) + 16 2}{2 ( - 32 )}, u_{2} = \frac{- ( - 16 6 ) - 16 2}{2 ( - 32 )}

u = \frac{- ( - 16 6 ) + 16 2}{2 ( - 32 )} : - \frac{6 + 2}{4}

\frac{- ( - 16 6 ) + 16 2}{2 ( - 32 )}

去除括号:

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

= \frac{16 6 + 16 2}{- 2 \cdot 32}

数字相乘:

2 \cdot 32 = 64

= \frac{16 6 + 16 2}{- 64}

使用分式法则:

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

= - \frac{16 6 + 16 2}{64}

消掉

\frac{16 6 + 16 2}{64} : \frac{6 + 2}{4}

\frac{16 6 + 16 2}{64}

因式分解出通项

16

= \frac{16 ( 6 + 2 )}{64}

约分:

16

= \frac{6 + 2}{4}

= - \frac{6 + 2}{4}

u = \frac{- ( - 16 6 ) - 16 2}{2 ( - 32 )} : - \frac{6 - 2}{4}

\frac{- ( - 16 6 ) - 16 2}{2 ( - 32 )}

去除括号:

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

= \frac{16 6 - 16 2}{- 2 \cdot 32}

数字相乘:

2 \cdot 32 = 64

= \frac{16 6 - 16 2}{- 64}

使用分式法则:

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

= - \frac{16 6 - 16 2}{64}

消掉

\frac{16 6 - 16 2}{64} : \frac{6 - 2}{4}

\frac{16 6 - 16 2}{64}

因式分解出通项

16

= \frac{16 ( 6 - 2 )}{64}

约分:

16

= \frac{6 - 2}{4}

= - \frac{6 - 2}{4}

二次方程组的解是:

u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

u = sin (x)

代回

sin (x) = - \frac{6 + 2}{4}, sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 + 2}{4}, sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 + 2}{4} : x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = π + arcsin (\frac{6 + 2}{4}) + 2 πn

sin (x) = - \frac{6 + 2}{4}

使用反三角函数性质

sin (x) = - \frac{6 + 2}{4}

sin (x) = - \frac{6 + 2}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

sin (x) = - \frac{6 - 2}{4} : x = arcsin (- \frac{6 - 2}{4}) + 2 πn, x = π + arcsin (\frac{6 - 2}{4}) + 2 πn

sin (x) = - \frac{6 - 2}{4}

使用反三角函数性质

sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 - 2}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

4 cos (x) - 4 sin (x) = 26

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{6 + 2}{4}) + 2 πn

的解:

真

arcsin (- \frac{6 + 2}{4}) + 2 πn

代入

n = 1

arcsin (- \frac{6 + 2}{4}) + 2 π 1

对于

4 cos (x) - 4 sin (x) = 26

代入

x = arcsin (- \frac{6 + 2}{4}) + 2 π 1

4 cos (arcsin (- \frac{6 + 2}{4}) + 2 π 1) - 4 sin (arcsin (- \frac{6 + 2}{4}) + 2 π 1) = 26

整理后得

4.89897 \dots = 4.89897 \dots

\Rightarrow 真

检验

π + arcsin (\frac{6 + 2}{4}) + 2 πn

的解:

假

π + arcsin (\frac{6 + 2}{4}) + 2 πn

代入

n = 1

π + arcsin (\frac{6 + 2}{4}) + 2 π 1

对于

4 cos (x) - 4 sin (x) = 26

代入

x = π + arcsin (\frac{6 + 2}{4}) + 2 π 1

4 cos (π + arcsin (\frac{6 + 2}{4}) + 2 π 1) - 4 sin (π + arcsin (\frac{6 + 2}{4}) + 2 π 1) = 26

整理后得

2.82842 \dots = 4.89897 \dots

\Rightarrow 假

检验

arcsin (- \frac{6 - 2}{4}) + 2 πn

的解:

真

arcsin (- \frac{6 - 2}{4}) + 2 πn

代入

n = 1

arcsin (- \frac{6 - 2}{4}) + 2 π 1

对于

4 cos (x) - 4 sin (x) = 26

代入

x = arcsin (- \frac{6 - 2}{4}) + 2 π 1

4 cos (arcsin (- \frac{6 - 2}{4}) + 2 π 1) - 4 sin (arcsin (- \frac{6 - 2}{4}) + 2 π 1) = 26

整理后得

4.89897 \dots = 4.89897 \dots

\Rightarrow 真

检验

π + arcsin (\frac{6 - 2}{4}) + 2 πn

的解:

假

π + arcsin (\frac{6 - 2}{4}) + 2 πn

代入

n = 1

π + arcsin (\frac{6 - 2}{4}) + 2 π 1

对于

4 cos (x) - 4 sin (x) = 26

代入

x = π + arcsin (\frac{6 - 2}{4}) + 2 π 1

4 cos (π + arcsin (\frac{6 - 2}{4}) + 2 π 1) - 4 sin (π + arcsin (\frac{6 - 2}{4}) + 2 π 1) = 26

整理后得

- 2.82842 \dots = 4.89897 \dots

\Rightarrow 假

x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = arcsin (- \frac{6 - 2}{4}) + 2 πn

以小数形式表示解

x = - 1.30899 \dots + 2 πn, x = - 0.26179 \dots + 2 πn

作图

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