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受欢迎的三角函数 >

3cos(2θ)-3cos(θ)+3=4cos(θ)+10

解答

3 cos (2 θ) - 3 cos (θ) + 3 = 4 cos (θ) + 10

解答

θ = 2.55590 \dots + 2 πn, θ = - 2.55590 \dots + 2 πn

+1

度数

θ = 146.44269 \dots^{\circ} + 36 0^{\circ} n, θ = - 146.44269 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 θ) - 3 cos (θ) + 3 = 4 cos (θ) + 10

两边减去

4 cos (θ) + 10

3 cos (2 θ) - 7 cos (θ) - 7 = 0

使用三角恒等式改写

- 7 + 3 cos (2 θ) - 7 cos (θ)

使用倍角公式:

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

= - 7 + 3 (2 cos^{2} (θ) - 1) - 7 cos (θ)

化简

- 7 + 3 (2 cos^{2} (θ) - 1) - 7 cos (θ) : 6 cos^{2} (θ) - 7 cos (θ) - 10

- 7 + 3 (2 cos^{2} (θ) - 1) - 7 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

= - 7 + 6 cos^{2} (θ) - 3 - 7 cos (θ)

化简

- 7 + 6 cos^{2} (θ) - 3 - 7 cos (θ) : 6 cos^{2} (θ) - 7 cos (θ) - 10

- 7 + 6 cos^{2} (θ) - 3 - 7 cos (θ)

对同类项分组

= 6 cos^{2} (θ) - 7 cos (θ) - 7 - 3

数字相减:

- 7 - 3 = - 10

= 6 cos^{2} (θ) - 7 cos (θ) - 10

= 6 cos^{2} (θ) - 7 cos (θ) - 10

= 6 cos^{2} (θ) - 7 cos (θ) - 10

- 10 + 6 cos^{2} (θ) - 7 cos (θ) = 0

用替代法求解

- 10 + 6 cos^{2} (θ) - 7 cos (θ) = 0

令

: cos (θ) = u

- 10 + 6 u^{2} - 7 u = 0

- 10 + 6 u^{2} - 7 u = 0 : u = 2, u = - \frac{5}{6}

- 10 + 6 u^{2} - 7 u = 0

改写成标准形式

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

6 u^{2} - 7 u - 10 = 0

使用求根公式求解

6 u^{2} - 7 u - 10 = 0

二次方程求根公式:

若

a = 6, b = - 7, c = - 10

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 10 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 10 )}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 (- 10) = 17

(- 7)^{2} - 4 \cdot 6 (- 10)

使用法则

- (- a) = a

= (- 7)^{2} + 4 \cdot 6 \cdot 10

使用指数法则:

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

若

n

是偶数

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

= 7^{2} + 4 \cdot 6 \cdot 10

数字相乘:

4 \cdot 6 \cdot 10 = 240

= 7^{2} + 240

7^{2} = 49

= 49 + 240

数字相加:

49 + 240 = 289

= 289

因式分解数字:

289 = 1 7^{2}

= 1 7^{2}

使用根式运算法则:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- ( - 7 ) \pm 17}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 17}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 17}{2 \cdot 6}

u = \frac{- ( - 7 ) + 17}{2 \cdot 6} : 2

\frac{- ( - 7 ) + 17}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 + 17}{2 \cdot 6}

数字相加:

7 + 17 = 24

= \frac{24}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{24}{12}

数字相除:

\frac{24}{12} = 2

= 2

u = \frac{- ( - 7 ) - 17}{2 \cdot 6} : - \frac{5}{6}

\frac{- ( - 7 ) - 17}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 - 17}{2 \cdot 6}

数字相减:

7 - 17 = - 10

= \frac{- 10}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 10}{12}

使用分式法则:

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

= - \frac{10}{12}

约分:

2

= - \frac{5}{6}

二次方程组的解是:

u = 2, u = - \frac{5}{6}

u = cos (θ)

代回

cos (θ) = 2, cos (θ) = - \frac{5}{6}

cos (θ) = 2, cos (θ) = - \frac{5}{6}

cos (θ) = 2 :

无解

cos (θ) = 2

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - \frac{5}{6} : θ = arccos (- \frac{5}{6}) + 2 πn, θ = - arccos (- \frac{5}{6}) + 2 πn

cos (θ) = - \frac{5}{6}

使用反三角函数性质

cos (θ) = - \frac{5}{6}

cos (θ) = - \frac{5}{6}

的通解

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

θ = arccos (- \frac{5}{6}) + 2 πn, θ = - arccos (- \frac{5}{6}) + 2 πn

θ = arccos (- \frac{5}{6}) + 2 πn, θ = - arccos (- \frac{5}{6}) + 2 πn

合并所有解

θ = arccos (- \frac{5}{6}) + 2 πn, θ = - arccos (- \frac{5}{6}) + 2 πn

以小数形式表示解

θ = 2.55590 \dots + 2 πn, θ = - 2.55590 \dots + 2 πn

作图

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