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受欢迎的三角函数 >

-3cos(2θ)+cos(θ)-10=-6cos(θ)-5

解答

- 3 cos (2 θ) + cos (θ) - 10 = - 6 cos (θ) - 5

解答

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 3 cos (2 θ) + cos (θ) - 10 = - 6 cos (θ) - 5

两边减去

- 6 cos (θ) - 5

7 cos (θ) - 3 cos (2 θ) - 5 = 0

使用三角恒等式改写

- 5 - 3 cos (2 θ) + 7 cos (θ)

使用倍角公式:

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

= - 5 - 3 (2 cos^{2} (θ) - 1) + 7 cos (θ)

化简

- 5 - 3 (2 cos^{2} (θ) - 1) + 7 cos (θ) : 7 cos (θ) - 6 cos^{2} (θ) - 2

- 5 - 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

使用加减运算法则

- (- a) = a

= - 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

= - 5 - 6 cos^{2} (θ) + 3 + 7 cos (θ)

化简

- 5 - 6 cos^{2} (θ) + 3 + 7 cos (θ) : 7 cos (θ) - 6 cos^{2} (θ) - 2

- 5 - 6 cos^{2} (θ) + 3 + 7 cos (θ)

对同类项分组

= - 6 cos^{2} (θ) + 7 cos (θ) - 5 + 3

数字相加/相减:

- 5 + 3 = - 2

= 7 cos (θ) - 6 cos^{2} (θ) - 2

= 7 cos (θ) - 6 cos^{2} (θ) - 2

= 7 cos (θ) - 6 cos^{2} (θ) - 2

- 2 - 6 cos^{2} (θ) + 7 cos (θ) = 0

用替代法求解

- 2 - 6 cos^{2} (θ) + 7 cos (θ) = 0

令

: cos (θ) = u

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

- 2 - 6 u^{2} + 7 u = 0 : u = \frac{1}{2}, u = \frac{2}{3}

- 2 - 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 - 2 = 0

使用求根公式求解

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

二次方程求根公式:

若

a = - 6, b = 7, c = - 2

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

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

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

7^{2} - 4 (- 6) (- 2)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 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 ( - 6 )}

将解分隔开

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

u = \frac{- 7 + 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 7 + 1}{2 ( - 6 )}

去除括号:

(- a) = - a

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

数字相加/相减:

- 7 + 1 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u = \frac{- 7 - 1}{2 ( - 6 )} : \frac{2}{3}

\frac{- 7 - 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 7 - 1}{- 2 \cdot 6}

数字相减:

- 7 - 1 = - 8

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

使用分式法则:

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

= \frac{8}{12}

约分:

4

= \frac{2}{3}

二次方程组的解是:

u = \frac{1}{2}, u = \frac{2}{3}

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = \frac{2}{3}

cos (θ) = \frac{1}{2}, cos (θ) = \frac{2}{3}

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

cos (θ) = \frac{1}{2}

cos (θ) = \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}

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

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

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

cos (θ) = \frac{2}{3}

使用反三角函数性质

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

作图

流行的例子

7cot(θ)=3csc(θ)

7 cot (θ) = 3 csc (θ)

5 sin (x) = 2

4(cot(θ)+1)=2(cot(θ)+2)

4 (cot (θ) + 1) = 2 (cot (θ) + 2)

sin (3 t) = 0

sin^2(x)-3cos(x)-3=0

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