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受欢迎的三角函数 >

4cos^2(x)-2cos(x)-3=0

解答

4 cos^{2} (x) - 2 cos (x) - 3 = 0

解答

x = 2.28020 \dots + 2 πn, x = - 2.28020 \dots + 2 πn

+1

度数

x = 130.64631 \dots^{\circ} + 36 0^{\circ} n, x = - 130.64631 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos^{2} (x) - 2 cos (x) - 3 = 0

用替代法求解

4 cos^{2} (x) - 2 cos (x) - 3 = 0

令

: cos (x) = u

4 u^{2} - 2 u - 3 = 0

4 u^{2} - 2 u - 3 = 0 : u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

4 u^{2} - 2 u - 3 = 0

使用求根公式求解

4 u^{2} - 2 u - 3 = 0

二次方程求根公式:

若

a = 4, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 3) = 213

(- 2)^{2} - 4 \cdot 4 (- 3)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 4 \cdot 3

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 4 \cdot 3 = 48

= 2^{2} + 48

2^{2} = 4

= 4 + 48

数字相加:

4 + 48 = 52

= 52

52

质因数分解:

2^{2} \cdot 13

52

52

除以

252 = 26 \cdot 2

= 2 \cdot 26

26

除以

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 13

2, 13

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 13

= 2^{2} \cdot 13

= 2^{2} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 213

u_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2 13}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 13}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 13}{2 \cdot 4} : \frac{1 + 13}{4}

\frac{- ( - 2 ) + 2 13}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 + 2 13}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 + 2 13}{8}

分解

2 + 213 : 2 (1 + 13)

2 + 213

改写为

= 2 \cdot 1 + 213

因式分解出通项

2

= 2 (1 + 13)

= \frac{2 ( 1 + 13 )}{8}

约分:

2

= \frac{1 + 13}{4}

u = \frac{- ( - 2 ) - 2 13}{2 \cdot 4} : \frac{1 - 13}{4}

\frac{- ( - 2 ) - 2 13}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 - 2 13}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 - 2 13}{8}

分解

2 - 213 : 2 (1 - 13)

2 - 213

改写为

= 2 \cdot 1 - 213

因式分解出通项

2

= 2 (1 - 13)

= \frac{2 ( 1 - 13 )}{8}

约分:

2

= \frac{1 - 13}{4}

二次方程组的解是:

u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

u = cos (x)

代回

cos (x) = \frac{1 + 13}{4}, cos (x) = \frac{1 - 13}{4}

cos (x) = \frac{1 + 13}{4}, cos (x) = \frac{1 - 13}{4}

cos (x) = \frac{1 + 13}{4} :

无解

cos (x) = \frac{1 + 13}{4}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{1 - 13}{4} : x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

cos (x) = \frac{1 - 13}{4}

使用反三角函数性质

cos (x) = \frac{1 - 13}{4}

cos (x) = \frac{1 - 13}{4}

的通解

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

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

合并所有解

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

以小数形式表示解

x = 2.28020 \dots + 2 πn, x = - 2.28020 \dots + 2 πn

作图

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