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受欢迎的三角函数 >

3cos(2θ)+5cos(θ)+2=0

解答

3 cos (2 θ) + 5 cos (θ) + 2 = 0

解答

θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn, θ = π + 2 πn

+1

度数

θ = 80.40593 \dots^{\circ} + 36 0^{\circ} n, θ = 279.59406 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 θ) + 5 cos (θ) + 2 = 0

使用三角恒等式改写

2 + 3 cos (2 θ) + 5 cos (θ)

使用倍角公式:

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

= 2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ)

化简

2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) - 1

2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ)

乘开

3 (2 cos^{2} (θ) - 1) : 6 cos^{2} (θ) - 3

3 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (θ), c = 1

= 3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

化简

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1 : 6 cos^{2} (θ) - 3

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

数字相乘:

3 \cdot 2 = 6

= 6 cos^{2} (θ) - 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 6 cos^{2} (θ) - 3

= 6 cos^{2} (θ) - 3

= 2 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

化简

2 + 6 cos^{2} (θ) - 3 + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) - 1

2 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

对同类项分组

= 6 cos^{2} (θ) + 5 cos (θ) + 2 - 3

数字相加/相减:

2 - 3 = - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

- 1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

用替代法求解

- 1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

令

: cos (θ) = u

- 1 + 5 u + 6 u^{2} = 0

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

- 1 + 5 u + 6 u^{2} = 0

改写成标准形式

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

6 u^{2} + 5 u - 1 = 0

使用求根公式求解

6 u^{2} + 5 u - 1 = 0

二次方程求根公式:

若

a = 6, b = 5, c = - 1

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

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

5^{2} - 4 \cdot 6 (- 1) = 7

5^{2} - 4 \cdot 6 (- 1)

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 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{- 5 \pm 7}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 \cdot 6}, u_{2} = \frac{- 5 - 7}{2 \cdot 6}

u = \frac{- 5 + 7}{2 \cdot 6} : \frac{1}{6}

\frac{- 5 + 7}{2 \cdot 6}

数字相加/相减:

- 5 + 7 = 2

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

数字相乘:

2 \cdot 6 = 12

= \frac{2}{12}

约分:

2

= \frac{1}{6}

u = \frac{- 5 - 7}{2 \cdot 6} : - 1

\frac{- 5 - 7}{2 \cdot 6}

数字相减:

- 5 - 7 = - 12

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 12}{12}

使用分式法则:

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

= - \frac{12}{12}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{1}{6}, u = - 1

u = cos (θ)

代回

cos (θ) = \frac{1}{6}, cos (θ) = - 1

cos (θ) = \frac{1}{6}, cos (θ) = - 1

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

cos (θ) = \frac{1}{6}

使用反三角函数性质

cos (θ) = \frac{1}{6}

cos (θ) = \frac{1}{6}

的通解

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

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

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

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

合并所有解

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

以小数形式表示解

θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn, θ = π + 2 πn

作图

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