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^2(x/2)-cos(x/2)=0

解答

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

解答

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

+1

度数

x = 77.33656 \dots^{\circ} + 72 0^{\circ} n, x = 642.66343 \dots^{\circ} + 72 0^{\circ} n

求解步骤

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

使用三角恒等式改写

- cos (\frac{x}{2}) + 2 sin^{2} (\frac{x}{2})

使用毕达哥拉斯恒等式:

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

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

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

- cos (\frac{x}{2}) + (1 - cos^{2} (\frac{x}{2})) \cdot 2 = 0

用替代法求解

- cos (\frac{x}{2}) + (1 - cos^{2} (\frac{x}{2})) \cdot 2 = 0

令

: cos (\frac{x}{2}) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

展开

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

= - u + 2 (1 - u^{2})

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= - u + 2 - 2 u^{2}

- u + 2 - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} - u + 2 = 0

使用求根公式求解

- 2 u^{2} - u + 2 = 0

二次方程求根公式:

若

a = - 2, b = - 1, c = 2

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

(- 1)^{2} - 4 (- 2) \cdot 2

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

使用分式法则:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

使用分式法则:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

u = cos (\frac{x}{2})

代回

cos (\frac{x}{2}) = - \frac{1 + 17}{4}, cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = - \frac{1 + 17}{4}, cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = - \frac{1 + 17}{4} :

无解

cos (\frac{x}{2}) = - \frac{1 + 17}{4}

- 1 \leq cos (x) \leq 1

无解

cos (\frac{x}{2}) = \frac{17 - 1}{4} : x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

cos (\frac{x}{2}) = \frac{17 - 1}{4}

使用反三角函数性质

cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = \frac{17 - 1}{4}

的通解

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

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

解

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn : x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn : 2 arccos (\frac{17 - 1}{4}) + 4 πn

2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

解

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn : x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn : 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

合并所有解

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

以小数形式表示解

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

作图

流行的例子

cos(pi/4-x)= 12/13

solvefor x,y=fcos(x)

sqrt(3)cot(2x)=-1,0<= x<= pi

5sin^2(x)-cos(x)=2

(cot(θ)-1)(2sin(θ)+sqrt(3))=0