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受欢迎的三角函数 >

5cos(x)=4sin(x)+4

解答

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

解答

x = \frac{3 π}{2} + 2 πn, x = 0.22131 \dots + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 12.68038 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

- 16 - 16 sin^{2} (x) + 25 cos^{2} (x) - 32 sin (x)

使用毕达哥拉斯恒等式:

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

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

= - 16 - 16 sin^{2} (x) + 25 (1 - sin^{2} (x)) - 32 sin (x)

化简

- 16 - 16 sin^{2} (x) + 25 (1 - sin^{2} (x)) - 32 sin (x) : - 41 sin^{2} (x) - 32 sin (x) + 9

- 16 - 16 sin^{2} (x) + 25 (1 - sin^{2} (x)) - 32 sin (x)

乘开

25 (1 - sin^{2} (x)) : 25 - 25 sin^{2} (x)

25 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 25, b = 1, c = sin^{2} (x)

= 25 \cdot 1 - 25 sin^{2} (x)

数字相乘:

25 \cdot 1 = 25

= 25 - 25 sin^{2} (x)

= - 16 - 16 sin^{2} (x) + 25 - 25 sin^{2} (x) - 32 sin (x)

化简

- 16 - 16 sin^{2} (x) + 25 - 25 sin^{2} (x) - 32 sin (x) : - 41 sin^{2} (x) - 32 sin (x) + 9

- 16 - 16 sin^{2} (x) + 25 - 25 sin^{2} (x) - 32 sin (x)

对同类项分组

= - 16 sin^{2} (x) - 25 sin^{2} (x) - 32 sin (x) - 16 + 25

同类项相加:

- 16 sin^{2} (x) - 25 sin^{2} (x) = - 41 sin^{2} (x)

= - 41 sin^{2} (x) - 32 sin (x) - 16 + 25

数字相加/相减:

- 16 + 25 = 9

= - 41 sin^{2} (x) - 32 sin (x) + 9

= - 41 sin^{2} (x) - 32 sin (x) + 9

= - 41 sin^{2} (x) - 32 sin (x) + 9

9 - 32 sin (x) - 41 sin^{2} (x) = 0

用替代法求解

9 - 32 sin (x) - 41 sin^{2} (x) = 0

令

: sin (x) = u

9 - 32 u - 41 u^{2} = 0

9 - 32 u - 41 u^{2} = 0 : u = - 1, u = \frac{9}{41}

9 - 32 u - 41 u^{2} = 0

改写成标准形式

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

- 41 u^{2} - 32 u + 9 = 0

使用求根公式求解

- 41 u^{2} - 32 u + 9 = 0

二次方程求根公式:

若

a = - 41, b = - 32, c = 9

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 41 ) \cdot 9}{2 ( - 41 )}

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 41 ) \cdot 9}{2 ( - 41 )}

(- 32)^{2} - 4 (- 41) \cdot 9 = 50

(- 32)^{2} - 4 (- 41) \cdot 9

使用法则

- (- a) = a

= (- 32)^{2} + 4 \cdot 41 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 32)^{2} = 3 2^{2}

= 3 2^{2} + 4 \cdot 41 \cdot 9

数字相乘:

4 \cdot 41 \cdot 9 = 1476

= 3 2^{2} + 1476

3 2^{2} = 1024

= 1024 + 1476

数字相加:

1024 + 1476 = 2500

= 2500

因式分解数字:

2500 = 5 0^{2}

= 5 0^{2}

使用根式运算法则:

n a^{n} = a

5 0^{2} = 50

= 50

u_{1, 2} = \frac{- ( - 32 ) \pm 50}{2 ( - 41 )}

将解分隔开

u_{1} = \frac{- ( - 32 ) + 50}{2 ( - 41 )}, u_{2} = \frac{- ( - 32 ) - 50}{2 ( - 41 )}

u = \frac{- ( - 32 ) + 50}{2 ( - 41 )} : - 1

\frac{- ( - 32 ) + 50}{2 ( - 41 )}

去除括号:

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

= \frac{32 + 50}{- 2 \cdot 41}

数字相加:

32 + 50 = 82

= \frac{82}{- 2 \cdot 41}

数字相乘:

2 \cdot 41 = 82

= \frac{82}{- 82}

使用分式法则:

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

= - \frac{82}{82}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 32 ) - 50}{2 ( - 41 )} : \frac{9}{41}

\frac{- ( - 32 ) - 50}{2 ( - 41 )}

去除括号:

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

= \frac{32 - 50}{- 2 \cdot 41}

数字相减:

32 - 50 = - 18

= \frac{- 18}{- 2 \cdot 41}

数字相乘:

2 \cdot 41 = 82

= \frac{- 18}{- 82}

使用分式法则:

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

= \frac{18}{82}

约分:

2

= \frac{9}{41}

二次方程组的解是:

u = - 1, u = \frac{9}{41}

u = sin (x)

代回

sin (x) = - 1, sin (x) = \frac{9}{41}

sin (x) = - 1, sin (x) = \frac{9}{41}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 1

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{9}{41} : x = arcsin (\frac{9}{41}) + 2 πn, x = π - arcsin (\frac{9}{41}) + 2 πn

sin (x) = \frac{9}{41}

使用反三角函数性质

sin (x) = \frac{9}{41}

sin (x) = \frac{9}{41}

的通解

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

x = arcsin (\frac{9}{41}) + 2 πn, x = π - arcsin (\frac{9}{41}) + 2 πn

x = arcsin (\frac{9}{41}) + 2 πn, x = π - arcsin (\frac{9}{41}) + 2 πn

合并所有解

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{9}{41}) + 2 πn, x = π - arcsin (\frac{9}{41}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{3 π}{2} + 2 πn

的解:

真

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

5 cos (\frac{3 π}{2} + 2 π 1) = 4 sin (\frac{3 π}{2} + 2 π 1) + 4

整理后得

0 = 0

\Rightarrow 真

检验

arcsin (\frac{9}{41}) + 2 πn

的解:

真

arcsin (\frac{9}{41}) + 2 πn

代入

n = 1

arcsin (\frac{9}{41}) + 2 π 1

对于

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

代入

x = arcsin (\frac{9}{41}) + 2 π 1

5 cos (arcsin (\frac{9}{41}) + 2 π 1) = 4 sin (arcsin (\frac{9}{41}) + 2 π 1) + 4

整理后得

4.87804 \dots = 4.87804 \dots

\Rightarrow 真

检验

π - arcsin (\frac{9}{41}) + 2 πn

的解:

假

π - arcsin (\frac{9}{41}) + 2 πn

代入

n = 1

π - arcsin (\frac{9}{41}) + 2 π 1

对于

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

代入

x = π - arcsin (\frac{9}{41}) + 2 π 1

5 cos (π - arcsin (\frac{9}{41}) + 2 π 1) = 4 sin (π - arcsin (\frac{9}{41}) + 2 π 1) + 4

整理后得

- 4.87804 \dots = 4.87804 \dots

\Rightarrow 假

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{9}{41}) + 2 πn

以小数形式表示解

x = \frac{3 π}{2} + 2 πn, x = 0.22131 \dots + 2 πn

作图

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