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受欢迎的三角函数 >

-5cos^2(θ)-4cos(θ)+4=0

解答

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

解答

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

+1

度数

θ = 54.56381 \dots^{\circ} + 36 0^{\circ} n, θ = 305.43618 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

用替代法求解

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

令

: cos (θ) = u

- 5 u^{2} - 4 u + 4 = 0

- 5 u^{2} - 4 u + 4 = 0 : u = - \frac{2 ( 1 + 6 )}{5}, u = \frac{2 ( 6 - 1 )}{5}

- 5 u^{2} - 4 u + 4 = 0

使用求根公式求解

- 5 u^{2} - 4 u + 4 = 0

二次方程求根公式:

若

a = - 5, b = - 4, c = 4

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 4}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 4}{2 ( - 5 )}

(- 4)^{2} - 4 (- 5) \cdot 4 = 46

(- 4)^{2} - 4 (- 5) \cdot 4

使用法则

- (- a) = a

= (- 4)^{2} + 4 \cdot 5 \cdot 4

使用指数法则:

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

若

n

是偶数

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

= 4^{2} + 4 \cdot 5 \cdot 4

数字相乘:

4 \cdot 5 \cdot 4 = 80

= 4^{2} + 80

4^{2} = 16

= 16 + 80

数字相加:

16 + 80 = 96

= 96

96

质因数分解:

2^{5} \cdot 3

96

96

除以

296 = 48 \cdot 2

= 2 \cdot 48

48

除以

248 = 24 \cdot 2

= 2 \cdot 2 \cdot 24

24

除以

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{5} \cdot 3

= 2^{5} \cdot 3

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{4} 2 \cdot 3

使用根式运算法则:

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

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

= 2^{2} 2 \cdot 3

整理后得

= 46

u_{1, 2} = \frac{- ( - 4 ) \pm 4 6}{2 ( - 5 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 4 6}{2 ( - 5 )}, u_{2} = \frac{- ( - 4 ) - 4 6}{2 ( - 5 )}

u = \frac{- ( - 4 ) + 4 6}{2 ( - 5 )} : - \frac{2 ( 1 + 6 )}{5}

\frac{- ( - 4 ) + 4 6}{2 ( - 5 )}

去除括号:

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

= \frac{4 + 4 6}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{4 + 4 6}{- 10}

使用分式法则:

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

= - \frac{4 + 4 6}{10}

消掉

\frac{4 + 4 6}{10} : \frac{2 ( 1 + 6 )}{5}

\frac{4 + 4 6}{10}

分解

4 + 46 : 4 (1 + 6)

4 + 46

改写为

= 4 \cdot 1 + 46

因式分解出通项

4

= 4 (1 + 6)

= \frac{4 ( 1 + 6 )}{10}

约分:

2

= \frac{2 ( 1 + 6 )}{5}

= - \frac{2 ( 1 + 6 )}{5}

u = \frac{- ( - 4 ) - 4 6}{2 ( - 5 )} : \frac{2 ( 6 - 1 )}{5}

\frac{- ( - 4 ) - 4 6}{2 ( - 5 )}

去除括号:

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

= \frac{4 - 4 6}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{4 - 4 6}{- 10}

使用分式法则:

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

4 - 46 = - (46 - 4)

= \frac{4 6 - 4}{10}

分解

46 - 4 : 4 (6 - 1)

46 - 4

改写为

= 46 - 4 \cdot 1

因式分解出通项

4

= 4 (6 - 1)

= \frac{4 ( 6 - 1 )}{10}

约分:

2

= \frac{2 ( 6 - 1 )}{5}

二次方程组的解是:

u = - \frac{2 ( 1 + 6 )}{5}, u = \frac{2 ( 6 - 1 )}{5}

u = cos (θ)

代回

cos (θ) = - \frac{2 ( 1 + 6 )}{5}, cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = - \frac{2 ( 1 + 6 )}{5}, cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = - \frac{2 ( 1 + 6 )}{5} :

无解

cos (θ) = - \frac{2 ( 1 + 6 )}{5}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = \frac{2 ( 6 - 1 )}{5} : θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

cos (θ) = \frac{2 ( 6 - 1 )}{5}

使用反三角函数性质

cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = \frac{2 ( 6 - 1 )}{5}

的通解

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

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

合并所有解

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

以小数形式表示解

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

作图

流行的例子

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sin(θ)sec(θ)=tan^2(θ)

sin (θ) sec (θ) = tan^{2} (θ)

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