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受欢迎的三角函数 >

8cos^2(x)-2cos(x)-1=0

解答

8 cos^{2} (x) - 2 cos (x) - 1 = 0

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

8 cos^{2} (x) - 2 cos (x) - 1 = 0

用替代法求解

8 cos^{2} (x) - 2 cos (x) - 1 = 0

令

: cos (x) = u

8 u^{2} - 2 u - 1 = 0

8 u^{2} - 2 u - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{4}

8 u^{2} - 2 u - 1 = 0

使用求根公式求解

8 u^{2} - 2 u - 1 = 0

二次方程求根公式:

若

a = 8, b = - 2, c = - 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 1) = 6

(- 2)^{2} - 4 \cdot 8 (- 1)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 8 \cdot 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 8}

u = \frac{- ( - 2 ) + 6}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 2 ) + 6}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{2 + 6}{2 \cdot 8}

数字相加:

2 + 6 = 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

u = \frac{- ( - 2 ) - 6}{2 \cdot 8} : - \frac{1}{4}

\frac{- ( - 2 ) - 6}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{2 - 6}{2 \cdot 8}

数字相减:

2 - 6 = - 4

= \frac{- 4}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4}{16}

使用分式法则:

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

= - \frac{4}{16}

约分:

4

= - \frac{1}{4}

二次方程组的解是:

u = \frac{1}{2}, u = - \frac{1}{4}

u = cos (x)

代回

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

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

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

以小数形式表示解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

作图

流行的例子

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