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 >

2sin(x)-cos(x)=0.5

解答

2 sin (x) - cos (x) = 0.5

解答

x = - 2.90345 \dots + 2 πn, x = 0.68916 \dots + 2 πn

+1

度数

x = - 166.35591 \dots^{\circ} + 36 0^{\circ} n, x = 39.48601 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x) - cos (x) = 0.5

两边加上

cos (x)

2 sin (x) = 0.5 + cos (x)

两边进行平方

(2 sin (x))^{2} = (0.5 + cos (x))^{2}

两边减去

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

4 sin^{2} (x) - 0.25 - cos (x) - cos^{2} (x) = 0

使用三角恒等式改写

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

= - 0.25 - cos (x) - cos^{2} (x) + 4 (1 - cos^{2} (x))

化简

- 0.25 - cos (x) - cos^{2} (x) + 4 (1 - cos^{2} (x)) : - 5 cos^{2} (x) - cos (x) + 3.75

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

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= - 0.25 - cos (x) - cos^{2} (x) + 4 - 4 cos^{2} (x)

化简

- 0.25 - cos (x) - cos^{2} (x) + 4 - 4 cos^{2} (x) : - 5 cos^{2} (x) - cos (x) + 3.75

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

对同类项分组

= - cos (x) - cos^{2} (x) - 4 cos^{2} (x) - 0.25 + 4

同类项相加:

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

= - cos (x) - 5 cos^{2} (x) - 0.25 + 4

数字相加/相减:

- 0.25 + 4 = 3.75

= - 5 cos^{2} (x) - cos (x) + 3.75

= - 5 cos^{2} (x) - cos (x) + 3.75

= - 5 cos^{2} (x) - cos (x) + 3.75

3.75 - cos (x) - 5 cos^{2} (x) = 0

用替代法求解

3.75 - cos (x) - 5 cos^{2} (x) = 0

令

: cos (x) = u

3.75 - u - 5 u^{2} = 0

3.75 - u - 5 u^{2} = 0 : u = - \frac{1 + 2 19}{10}, u = \frac{2 19 - 1}{10}

3.75 - u - 5 u^{2} = 0

在两边乘以

100

3.75 - u - 5 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

3.75 \cdot 100 - u \cdot 100 - 5 u^{2} \cdot 100 = 0 \cdot 100

整理后得

375 - 100 u - 500 u^{2} = 0

375 - 100 u - 500 u^{2} = 0

改写成标准形式

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

- 500 u^{2} - 100 u + 375 = 0

使用求根公式求解

- 500 u^{2} - 100 u + 375 = 0

二次方程求根公式:

若

a = - 500, b = - 100, c = 375

u_{1, 2} = \frac{- ( - 100 ) \pm ( - 100 ) ^{2} - 4 ( - 500 ) \cdot 375}{2 ( - 500 )}

u_{1, 2} = \frac{- ( - 100 ) \pm ( - 100 ) ^{2} - 4 ( - 500 ) \cdot 375}{2 ( - 500 )}

(- 100)^{2} - 4 (- 500) \cdot 375 = 20019

(- 100)^{2} - 4 (- 500) \cdot 375

使用法则

- (- a) = a

= (- 100)^{2} + 4 \cdot 500 \cdot 375

使用指数法则:

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

若

n

是偶数

(- 100)^{2} = 10 0^{2}

= 10 0^{2} + 4 \cdot 500 \cdot 375

数字相乘:

4 \cdot 500 \cdot 375 = 750000

= 10 0^{2} + 750000

10 0^{2} = 10000

= 10000 + 750000

数字相加:

10000 + 750000 = 760000

= 760000

760000

质因数分解:

2^{6} \cdot 5^{4} \cdot 19

760000

= 2^{6} \cdot 5^{4} \cdot 19

使用根式运算法则:

n ab = n a n b

= 19 2^{6} 5^{4}

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} 19 5^{4}

使用根式运算法则:

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

5^{4} = 5^{\frac{4}{2}} = 5^{2}

= 2^{3} \cdot 5^{2} 19

整理后得

= 20019

u_{1, 2} = \frac{- ( - 100 ) \pm 200 19}{2 ( - 500 )}

将解分隔开

u_{1} = \frac{- ( - 100 ) + 200 19}{2 ( - 500 )}, u_{2} = \frac{- ( - 100 ) - 200 19}{2 ( - 500 )}

u = \frac{- ( - 100 ) + 200 19}{2 ( - 500 )} : - \frac{1 + 2 19}{10}

\frac{- ( - 100 ) + 200 19}{2 ( - 500 )}

去除括号:

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

= \frac{100 + 200 19}{- 2 \cdot 500}

数字相乘:

2 \cdot 500 = 1000

= \frac{100 + 200 19}{- 1000}

使用分式法则:

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

= - \frac{100 + 200 19}{1000}

消掉

\frac{100 + 200 19}{1000} : \frac{1 + 2 19}{10}

\frac{100 + 200 19}{1000}

分解

100 + 20019 : 100 (1 + 219)

100 + 20019

改写为

= 100 \cdot 1 + 100 \cdot 219

因式分解出通项

100

= 100 (1 + 219)

= \frac{100 ( 1 + 2 19 )}{1000}

约分:

100

= \frac{1 + 2 19}{10}

= - \frac{1 + 2 19}{10}

u = \frac{- ( - 100 ) - 200 19}{2 ( - 500 )} : \frac{2 19 - 1}{10}

\frac{- ( - 100 ) - 200 19}{2 ( - 500 )}

去除括号:

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

= \frac{100 - 200 19}{- 2 \cdot 500}

数字相乘:

2 \cdot 500 = 1000

= \frac{100 - 200 19}{- 1000}

使用分式法则:

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

100 - 20019 = - (20019 - 100)

= \frac{200 19 - 100}{1000}

分解

20019 - 100 : 100 (219 - 1)

20019 - 100

改写为

= 100 \cdot 219 - 100 \cdot 1

因式分解出通项

100

= 100 (219 - 1)

= \frac{100 ( 2 19 - 1 )}{1000}

约分:

100

= \frac{2 19 - 1}{10}

二次方程组的解是:

u = - \frac{1 + 2 19}{10}, u = \frac{2 19 - 1}{10}

u = cos (x)

代回

cos (x) = - \frac{1 + 2 19}{10}, cos (x) = \frac{2 19 - 1}{10}

cos (x) = - \frac{1 + 2 19}{10}, cos (x) = \frac{2 19 - 1}{10}

cos (x) = - \frac{1 + 2 19}{10} : x = arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = - arccos (- \frac{1 + 2 19}{10}) + 2 πn

cos (x) = - \frac{1 + 2 19}{10}

使用反三角函数性质

cos (x) = - \frac{1 + 2 19}{10}

cos (x) = - \frac{1 + 2 19}{10}

的通解

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

x = arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = - arccos (- \frac{1 + 2 19}{10}) + 2 πn

x = arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = - arccos (- \frac{1 + 2 19}{10}) + 2 πn

cos (x) = \frac{2 19 - 1}{10} : x = arccos (\frac{2 19 - 1}{10}) + 2 πn, x = 2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

cos (x) = \frac{2 19 - 1}{10}

使用反三角函数性质

cos (x) = \frac{2 19 - 1}{10}

cos (x) = \frac{2 19 - 1}{10}

的通解

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

x = arccos (\frac{2 19 - 1}{10}) + 2 πn, x = 2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

x = arccos (\frac{2 19 - 1}{10}) + 2 πn, x = 2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

合并所有解

x = arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = - arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = arccos (\frac{2 19 - 1}{10}) + 2 πn, x = 2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

将解代入原方程进行验证

将它们代入

2 sin (x) - cos (x) = 0.5

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{1 + 2 19}{10}) + 2 πn

的解:

假

arccos (- \frac{1 + 2 19}{10}) + 2 πn

代入

n = 1

arccos (- \frac{1 + 2 19}{10}) + 2 π 1

对于

2 sin (x) - cos (x) = 0.5

代入

x = arccos (- \frac{1 + 2 19}{10}) + 2 π 1

2 sin (arccos (- \frac{1 + 2 19}{10}) + 2 π 1) - cos (arccos (- \frac{1 + 2 19}{10}) + 2 π 1) = 0.5

整理后得

1.44355 \dots = 0.5

\Rightarrow 假

检验

- arccos (- \frac{1 + 2 19}{10}) + 2 πn

的解:

真

- arccos (- \frac{1 + 2 19}{10}) + 2 πn

代入

n = 1

- arccos (- \frac{1 + 2 19}{10}) + 2 π 1

对于

2 sin (x) - cos (x) = 0.5

代入

x = - arccos (- \frac{1 + 2 19}{10}) + 2 π 1

2 sin (- arccos (- \frac{1 + 2 19}{10}) + 2 π 1) - cos (- arccos (- \frac{1 + 2 19}{10}) + 2 π 1) = 0.5

整理后得

0.5 = 0.5

\Rightarrow 真

检验

arccos (\frac{2 19 - 1}{10}) + 2 πn

的解:

真

arccos (\frac{2 19 - 1}{10}) + 2 πn

代入

n = 1

arccos (\frac{2 19 - 1}{10}) + 2 π 1

对于

2 sin (x) - cos (x) = 0.5

代入

x = arccos (\frac{2 19 - 1}{10}) + 2 π 1

2 sin (arccos (\frac{2 19 - 1}{10}) + 2 π 1) - cos (arccos (\frac{2 19 - 1}{10}) + 2 π 1) = 0.5

整理后得

0.5 = 0.5

\Rightarrow 真

检验

2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

的解:

假

2 π - arccos (\frac{2 19 - 1}{10}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 19 - 1}{10}) + 2 π 1

对于

2 sin (x) - cos (x) = 0.5

代入

x = 2 π - arccos (\frac{2 19 - 1}{10}) + 2 π 1

2 sin (2 π - arccos (\frac{2 19 - 1}{10}) + 2 π 1) - cos (2 π - arccos (\frac{2 19 - 1}{10}) + 2 π 1) = 0.5

整理后得

- 2.04355 \dots = 0.5

\Rightarrow 假

x = - arccos (- \frac{1 + 2 19}{10}) + 2 πn, x = arccos (\frac{2 19 - 1}{10}) + 2 πn

以小数形式表示解

x = - 2.90345 \dots + 2 πn, x = 0.68916 \dots + 2 πn

作图

流行的例子

cos(x)-tan(x)=0

cos (x) - tan (x) = 0

2sin(x)-cos(x)=0.2

2 sin (x) - cos (x) = 0.2

tan^2(x)-tan(x)=0,0<= x<= 2pi

tan^{2} (x) - tan (x) = 0, 0 \leq x \leq 2 π

cos (2 x) = \frac{1}{9}

((1))/((cot(θ)))=(-(7))/((25))

\frac{( 1 )}{( cot ( θ ) )} = \frac{- ( 7 )}{( 25 )}