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受欢迎的三角函数 >

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

解答

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

解答

x = - 2.76750 \dots + 2 πn, x = 0.55321 \dots + 2 πn

+1

度数

x = - 158.56649 \dots^{\circ} + 36 0^{\circ} n, x = 31.69659 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

cos (x)

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

两边进行平方

(2 sin (x))^{2} = (0.2 + cos (x))^{2}

两边减去

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

4 sin^{2} (x) - 0.04 - 0.4 cos (x) - cos^{2} (x) = 0

使用三角恒等式改写

- 0.04 - cos^{2} (x) - 0.4 cos (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)

= - 0.04 - cos^{2} (x) - 0.4 cos (x) + 4 (1 - cos^{2} (x))

化简

- 0.04 - cos^{2} (x) - 0.4 cos (x) + 4 (1 - cos^{2} (x)) : - 5 cos^{2} (x) - 0.4 cos (x) + 3.96

- 0.04 - cos^{2} (x) - 0.4 cos (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)

= - 0.04 - cos^{2} (x) - 0.4 cos (x) + 4 - 4 cos^{2} (x)

化简

- 0.04 - cos^{2} (x) - 0.4 cos (x) + 4 - 4 cos^{2} (x) : - 5 cos^{2} (x) - 0.4 cos (x) + 3.96

- 0.04 - cos^{2} (x) - 0.4 cos (x) + 4 - 4 cos^{2} (x)

对同类项分组

= - cos^{2} (x) - 0.4 cos (x) - 4 cos^{2} (x) - 0.04 + 4

同类项相加:

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

= - 5 cos^{2} (x) - 0.4 cos (x) - 0.04 + 4

数字相加/相减:

- 0.04 + 4 = 3.96

= - 5 cos^{2} (x) - 0.4 cos (x) + 3.96

= - 5 cos^{2} (x) - 0.4 cos (x) + 3.96

= - 5 cos^{2} (x) - 0.4 cos (x) + 3.96

3.96 - 0.4 cos (x) - 5 cos^{2} (x) = 0

用替代法求解

3.96 - 0.4 cos (x) - 5 cos^{2} (x) = 0

令

: cos (x) = u

3.96 - 0.4 u - 5 u^{2} = 0

3.96 - 0.4 u - 5 u^{2} = 0 : u = - \frac{1 + 4 31}{25}, u = \frac{4 31 - 1}{25}

3.96 - 0.4 u - 5 u^{2} = 0

在两边乘以

100

3.96 - 0.4 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

3.96 \cdot 100 - 0.4 u \cdot 100 - 5 u^{2} \cdot 100 = 0 \cdot 100

整理后得

396 - 40 u - 500 u^{2} = 0

396 - 40 u - 500 u^{2} = 0

改写成标准形式

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

- 500 u^{2} - 40 u + 396 = 0

使用求根公式求解

- 500 u^{2} - 40 u + 396 = 0

二次方程求根公式:

若

a = - 500, b = - 40, c = 396

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 500 ) \cdot 396}{2 ( - 500 )}

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 500 ) \cdot 396}{2 ( - 500 )}

(- 40)^{2} - 4 (- 500) \cdot 396 = 16031

(- 40)^{2} - 4 (- 500) \cdot 396

使用法则

- (- a) = a

= (- 40)^{2} + 4 \cdot 500 \cdot 396

使用指数法则:

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

若

n

是偶数

(- 40)^{2} = 4 0^{2}

= 4 0^{2} + 4 \cdot 500 \cdot 396

数字相乘:

4 \cdot 500 \cdot 396 = 792000

= 4 0^{2} + 792000

4 0^{2} = 1600

= 1600 + 792000

数字相加:

1600 + 792000 = 793600

= 793600

793600

质因数分解:

2^{10} \cdot 5^{2} \cdot 31

793600

= 2^{10} \cdot 5^{2} \cdot 31

使用根式运算法则:

n ab = n a n b

= 31 5^{2} 2^{10}

使用根式运算法则:

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

2^{10} = 2^{\frac{10}{2}} = 2^{5}

= 2^{5} 31 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 2^{5} \cdot 531

整理后得

= 16031

u_{1, 2} = \frac{- ( - 40 ) \pm 160 31}{2 ( - 500 )}

将解分隔开

u_{1} = \frac{- ( - 40 ) + 160 31}{2 ( - 500 )}, u_{2} = \frac{- ( - 40 ) - 160 31}{2 ( - 500 )}

u = \frac{- ( - 40 ) + 160 31}{2 ( - 500 )} : - \frac{1 + 4 31}{25}

\frac{- ( - 40 ) + 160 31}{2 ( - 500 )}

去除括号:

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

= \frac{40 + 160 31}{- 2 \cdot 500}

数字相乘:

2 \cdot 500 = 1000

= \frac{40 + 160 31}{- 1000}

使用分式法则:

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

= - \frac{40 + 160 31}{1000}

消掉

\frac{40 + 160 31}{1000} : \frac{1 + 4 31}{25}

\frac{40 + 160 31}{1000}

分解

40 + 16031 : 40 (1 + 431)

40 + 16031

改写为

= 40 \cdot 1 + 40 \cdot 431

因式分解出通项

40

= 40 (1 + 431)

= \frac{40 ( 1 + 4 31 )}{1000}

约分:

40

= \frac{1 + 4 31}{25}

= - \frac{1 + 4 31}{25}

u = \frac{- ( - 40 ) - 160 31}{2 ( - 500 )} : \frac{4 31 - 1}{25}

\frac{- ( - 40 ) - 160 31}{2 ( - 500 )}

去除括号:

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

= \frac{40 - 160 31}{- 2 \cdot 500}

数字相乘:

2 \cdot 500 = 1000

= \frac{40 - 160 31}{- 1000}

使用分式法则:

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

40 - 16031 = - (16031 - 40)

= \frac{160 31 - 40}{1000}

分解

16031 - 40 : 40 (431 - 1)

16031 - 40

改写为

= 40 \cdot 431 - 40 \cdot 1

因式分解出通项

40

= 40 (431 - 1)

= \frac{40 ( 4 31 - 1 )}{1000}

约分:

40

= \frac{4 31 - 1}{25}

二次方程组的解是:

u = - \frac{1 + 4 31}{25}, u = \frac{4 31 - 1}{25}

u = cos (x)

代回

cos (x) = - \frac{1 + 4 31}{25}, cos (x) = \frac{4 31 - 1}{25}

cos (x) = - \frac{1 + 4 31}{25}, cos (x) = \frac{4 31 - 1}{25}

cos (x) = - \frac{1 + 4 31}{25} : x = arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = - arccos (- \frac{1 + 4 31}{25}) + 2 πn

cos (x) = - \frac{1 + 4 31}{25}

使用反三角函数性质

cos (x) = - \frac{1 + 4 31}{25}

cos (x) = - \frac{1 + 4 31}{25}

的通解

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

x = arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = - arccos (- \frac{1 + 4 31}{25}) + 2 πn

x = arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = - arccos (- \frac{1 + 4 31}{25}) + 2 πn

cos (x) = \frac{4 31 - 1}{25} : x = arccos (\frac{4 31 - 1}{25}) + 2 πn, x = 2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

cos (x) = \frac{4 31 - 1}{25}

使用反三角函数性质

cos (x) = \frac{4 31 - 1}{25}

cos (x) = \frac{4 31 - 1}{25}

的通解

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

x = arccos (\frac{4 31 - 1}{25}) + 2 πn, x = 2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

x = arccos (\frac{4 31 - 1}{25}) + 2 πn, x = 2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

合并所有解

x = arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = - arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = arccos (\frac{4 31 - 1}{25}) + 2 πn, x = 2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{1 + 4 31}{25}) + 2 πn

的解:

假

arccos (- \frac{1 + 4 31}{25}) + 2 πn

代入

n = 1

arccos (- \frac{1 + 4 31}{25}) + 2 π 1

对于

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

代入

x = arccos (- \frac{1 + 4 31}{25}) + 2 π 1

2 sin (arccos (- \frac{1 + 4 31}{25}) + 2 π 1) - cos (arccos (- \frac{1 + 4 31}{25}) + 2 π 1) = 0.2

整理后得

1.66168 \dots = 0.2

\Rightarrow 假

检验

- arccos (- \frac{1 + 4 31}{25}) + 2 πn

的解:

真

- arccos (- \frac{1 + 4 31}{25}) + 2 πn

代入

n = 1

- arccos (- \frac{1 + 4 31}{25}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{1 + 4 31}{25}) + 2 π 1

2 sin (- arccos (- \frac{1 + 4 31}{25}) + 2 π 1) - cos (- arccos (- \frac{1 + 4 31}{25}) + 2 π 1) = 0.2

整理后得

0.2 = 0.2

\Rightarrow 真

检验

arccos (\frac{4 31 - 1}{25}) + 2 πn

的解:

真

arccos (\frac{4 31 - 1}{25}) + 2 πn

代入

n = 1

arccos (\frac{4 31 - 1}{25}) + 2 π 1

对于

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

代入

x = arccos (\frac{4 31 - 1}{25}) + 2 π 1

2 sin (arccos (\frac{4 31 - 1}{25}) + 2 π 1) - cos (arccos (\frac{4 31 - 1}{25}) + 2 π 1) = 0.2

整理后得

0.2 = 0.2

\Rightarrow 真

检验

2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

的解:

假

2 π - arccos (\frac{4 31 - 1}{25}) + 2 πn

代入

n = 1

2 π - arccos (\frac{4 31 - 1}{25}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{4 31 - 1}{25}) + 2 π 1

2 sin (2 π - arccos (\frac{4 31 - 1}{25}) + 2 π 1) - cos (2 π - arccos (\frac{4 31 - 1}{25}) + 2 π 1) = 0.2

整理后得

- 1.90168 \dots = 0.2

\Rightarrow 假

x = - arccos (- \frac{1 + 4 31}{25}) + 2 πn, x = arccos (\frac{4 31 - 1}{25}) + 2 πn

以小数形式表示解

x = - 2.76750 \dots + 2 πn, x = 0.55321 \dots + 2 πn

作图

流行的例子

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 )}

tan(θ)=(-1)/(-2)

tan (θ) = \frac{- 1}{- 2}

sin(θ)=-1/3 ,(3pi)/2 <θ<2pi

sin (θ) = - \frac{1}{3}, \frac{3 π}{2} < θ < 2 π