zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

sin(x)+4cos(x)+5=0

解答

sin (x) + 4 cos (x) + 5 = 0

解答

x \in R 无解

求解步骤

sin (x) + 4 cos (x) + 5 = 0

两边减去

4 cos (x)

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

两边进行平方

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

两边减去

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

(sin (x) + 5)^{2} - 16 cos^{2} (x) = 0

使用三角恒等式改写

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

= (5 + sin (x))^{2} - 16 (1 - sin^{2} (x))

化简

(5 + sin (x))^{2} - 16 (1 - sin^{2} (x)) : 17 sin^{2} (x) + 10 sin (x) + 9

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

(5 + sin (x))^{2} : 25 + 10 sin (x) + sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 5, b = sin (x)

= 5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x)

化简

5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x) : 25 + 10 sin (x) + sin^{2} (x)

5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x)

5^{2} = 25

= 25 + 2 \cdot 5 sin (x) + sin^{2} (x)

数字相乘:

2 \cdot 5 = 10

= 25 + 10 sin (x) + sin^{2} (x)

= 25 + 10 sin (x) + sin^{2} (x)

= 25 + 10 sin (x) + sin^{2} (x) - 16 (1 - sin^{2} (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)

使用加减运算法则

- (- a) = a

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

数字相乘:

16 \cdot 1 = 16

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

= 25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x)

化简

25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x) : 17 sin^{2} (x) + 10 sin (x) + 9

25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x)

对同类项分组

= 10 sin (x) + sin^{2} (x) + 16 sin^{2} (x) + 25 - 16

同类项相加:

sin^{2} (x) + 16 sin^{2} (x) = 17 sin^{2} (x)

= 10 sin (x) + 17 sin^{2} (x) + 25 - 16

数字相加/相减:

25 - 16 = 9

= 17 sin^{2} (x) + 10 sin (x) + 9

= 17 sin^{2} (x) + 10 sin (x) + 9

= 17 sin^{2} (x) + 10 sin (x) + 9

9 + 10 sin (x) + 17 sin^{2} (x) = 0

用替代法求解

9 + 10 sin (x) + 17 sin^{2} (x) = 0

令

: sin (x) = u

9 + 10 u + 17 u^{2} = 0

9 + 10 u + 17 u^{2} = 0 : u = - \frac{5}{17} + i \frac{8 2}{17}, u = - \frac{5}{17} - i \frac{8 2}{17}

9 + 10 u + 17 u^{2} = 0

改写成标准形式

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

17 u^{2} + 10 u + 9 = 0

使用求根公式求解

17 u^{2} + 10 u + 9 = 0

二次方程求根公式:

若

a = 17, b = 10, c = 9

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 17 \cdot 9}{2 \cdot 17}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 17 \cdot 9}{2 \cdot 17}

化简

1 0^{2} - 4 \cdot 17 \cdot 9 : 162 i

1 0^{2} - 4 \cdot 17 \cdot 9

数字相乘:

4 \cdot 17 \cdot 9 = 612

= 1 0^{2} - 612

使用虚数运算法则:

- a = i a

= i 612 - 1 0^{2}

- 1 0^{2} + 612 = 162

- 1 0^{2} + 612

1 0^{2} = 100

= - 100 + 612

数字相加/相减:

- 100 + 612 = 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

= 162 i

u_{1, 2} = \frac{- 10 \pm 16 2 i}{2 \cdot 17}

将解分隔开

u_{1} = \frac{- 10 + 16 2 i}{2 \cdot 17}, u_{2} = \frac{- 10 - 16 2 i}{2 \cdot 17}

u = \frac{- 10 + 16 2 i}{2 \cdot 17} : - \frac{5}{17} + i \frac{8 2}{17}

\frac{- 10 + 16 2 i}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{- 10 + 16 2 i}{34}

分解

- 10 + 162 i : 2 (- 5 + 82 i)

- 10 + 162 i

改写为

= - 2 \cdot 5 + 2 \cdot 82 i

因式分解出通项

2

= 2 (- 5 + 82 i)

= \frac{2 ( - 5 + 8 2 i )}{34}

约分:

2

= \frac{- 5 + 8 2 i}{17}

将

\frac{- 5 + 8 2 i}{17}

改写成标准复数形式:

- \frac{5}{17} + \frac{8 2}{17} i

\frac{- 5 + 8 2 i}{17}

使用分式法则:

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

\frac{- 5 + 8 2 i}{17} = - \frac{5}{17} + \frac{8 2 i}{17}

= - \frac{5}{17} + \frac{8 2 i}{17}

= - \frac{5}{17} + \frac{8 2}{17} i

u = \frac{- 10 - 16 2 i}{2 \cdot 17} : - \frac{5}{17} - i \frac{8 2}{17}

\frac{- 10 - 16 2 i}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{- 10 - 16 2 i}{34}

分解

- 10 - 162 i : - 2 (5 + 82 i)

- 10 - 162 i

改写为

= - 2 \cdot 5 - 2 \cdot 82 i

因式分解出通项

2

= - 2 (5 + 82 i)

= - \frac{2 ( 5 + 8 2 i )}{34}

约分:

2

= - \frac{5 + 8 2 i}{17}

将

- \frac{5 + 8 2 i}{17}

改写成标准复数形式:

- \frac{5}{17} - \frac{8 2}{17} i

- \frac{5 + 8 2 i}{17}

使用分式法则:

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

\frac{5 + 8 2 i}{17} = - (\frac{5}{17}) - (\frac{8 2 i}{17})

= - (\frac{5}{17}) - (\frac{8 2 i}{17})

去除括号:

(a) = a

= - \frac{5}{17} - \frac{8 2 i}{17}

= - \frac{5}{17} - \frac{8 2}{17} i

二次方程组的解是:

u = - \frac{5}{17} + i \frac{8 2}{17}, u = - \frac{5}{17} - i \frac{8 2}{17}

u = sin (x)

代回

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}, sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}, sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

sin (x) = - \frac{5}{17} + i \frac{8 2}{17} :

无解

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}

无解

sin (x) = - \frac{5}{17} - i \frac{8 2}{17} :

无解

sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

无解

合并所有解

无解

将解代入原方程进行验证

将它们代入

sin (x) + 4 cos (x) + 5 = 0

检验解是否符合

去除与方程不符的解。

x \in R 无解

作图

流行的例子

sec (x) = - 5

sin(2t-pi/5)=-1/3

sin (2 t - \frac{π}{5}) = - \frac{1}{3}

2(sin(x))^2-cos(x)-1=0

2 (sin (x))^{2} - cos (x) - 1 = 0

arctan(3x)+arctan(x)= pi/4

arctan (3 x) + arctan (x) = \frac{π}{4}

4sin(θ)=sqrt(3)sec(θ),0<= θ<180

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}