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受欢迎的三角函数 >

12cos^2(x)-7cos(x)+1=0

解答

12 cos^{2} (x) - 7 cos (x) + 1 = 0

解答

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

+1

度数

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

12 cos^{2} (x) - 7 cos (x) + 1 = 0

用替代法求解

12 cos^{2} (x) - 7 cos (x) + 1 = 0

令

: cos (x) = u

12 u^{2} - 7 u + 1 = 0

12 u^{2} - 7 u + 1 = 0 : u = \frac{1}{3}, u = \frac{1}{4}

12 u^{2} - 7 u + 1 = 0

使用求根公式求解

12 u^{2} - 7 u + 1 = 0

二次方程求根公式:

若

a = 12, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 \cdot 1 = 1

(- 7)^{2} - 4 \cdot 12 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 12 \cdot 1

数字相乘:

4 \cdot 12 \cdot 1 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 12}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 12}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 12}

u = \frac{- ( - 7 ) + 1}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 7 ) + 1}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{7 + 1}{2 \cdot 12}

数字相加:

7 + 1 = 8

= \frac{8}{2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{8}{24}

约分:

8

= \frac{1}{3}

u = \frac{- ( - 7 ) - 1}{2 \cdot 12} : \frac{1}{4}

\frac{- ( - 7 ) - 1}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{7 - 1}{2 \cdot 12}

数字相减:

7 - 1 = 6

= \frac{6}{2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{6}{24}

约分:

6

= \frac{1}{4}

二次方程组的解是:

u = \frac{1}{3}, u = \frac{1}{4}

u = cos (x)

代回

cos (x) = \frac{1}{3}, cos (x) = \frac{1}{4}

cos (x) = \frac{1}{3}, cos (x) = \frac{1}{4}

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

使用反三角函数性质

cos (x) = \frac{1}{3}

cos (x) = \frac{1}{3}

的通解

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

以小数形式表示解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

作图

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