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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

- 1 - cos (x) + 2 cos (2 x)

使用倍角公式:

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

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

化简

- 1 - cos (x) + 2 (2 cos^{2} (x) - 1) : 4 cos^{2} (x) - cos (x) - 3

- 1 - cos (x) + 2 (2 cos^{2} (x) - 1)

乘开

2 (2 cos^{2} (x) - 1) : 4 cos^{2} (x) - 2

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (x), c = 1

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

化简

2 \cdot 2 cos^{2} (x) - 2 \cdot 1 : 4 cos^{2} (x) - 2

2 \cdot 2 cos^{2} (x) - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

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

数字相乘:

2 \cdot 1 = 2

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

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

= - 1 - cos (x) + 4 cos^{2} (x) - 2

化简

- 1 - cos (x) + 4 cos^{2} (x) - 2 : 4 cos^{2} (x) - cos (x) - 3

- 1 - cos (x) + 4 cos^{2} (x) - 2

对同类项分组

= - cos (x) + 4 cos^{2} (x) - 1 - 2

数字相减:

- 1 - 2 = - 3

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

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

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

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

用替代法求解

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

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = - 1, c = - 3

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

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

(- 1)^{2} - 4 \cdot 4 (- 3) = 7

(- 1)^{2} - 4 \cdot 4 (- 3)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \cdot 3

(- 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 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

数字相乘:

4 \cdot 4 \cdot 3 = 48

= 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

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

将解分隔开

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 4}

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

\frac{- ( - 1 ) + 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{1 + 7}{2 \cdot 4}

数字相加:

1 + 7 = 8

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

数字相乘:

2 \cdot 4 = 8

= \frac{8}{8}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 7}{2 \cdot 4} : - \frac{3}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{1 - 7}{2 \cdot 4}

数字相减:

1 - 7 = - 6

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 6}{8}

使用分式法则:

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

= - \frac{6}{8}

约分:

2

= - \frac{3}{4}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

cos (x) = - \frac{3}{4}

使用反三角函数性质

cos (x) = - \frac{3}{4}

cos (x) = - \frac{3}{4}

的通解

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

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

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

合并所有解

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

以小数形式表示解

x = 2 πn, x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

作图

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