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受欢迎的三角函数 >

6cos(2θ)=18cos(θ)-12

解答

6 cos (2 θ) = 18 cos (θ) - 12

解答

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

+1

度数

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

求解步骤

6 cos (2 θ) = 18 cos (θ) - 12

两边减去

18 cos (θ) - 12

6 cos (2 θ) - 18 cos (θ) + 12 = 0

使用三角恒等式改写

12 - 18 cos (θ) + 6 cos (2 θ)

使用倍角公式:

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

= 12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1)

化简

12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1) : 12 cos^{2} (θ) - 18 cos (θ) + 6

12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1)

乘开

6 (2 cos^{2} (θ) - 1) : 12 cos^{2} (θ) - 6

6 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 6, b = 2 cos^{2} (θ), c = 1

= 6 \cdot 2 cos^{2} (θ) - 6 \cdot 1

化简

6 \cdot 2 cos^{2} (θ) - 6 \cdot 1 : 12 cos^{2} (θ) - 6

6 \cdot 2 cos^{2} (θ) - 6 \cdot 1

数字相乘:

6 \cdot 2 = 12

= 12 cos^{2} (θ) - 6 \cdot 1

数字相乘:

6 \cdot 1 = 6

= 12 cos^{2} (θ) - 6

= 12 cos^{2} (θ) - 6

= 12 - 18 cos (θ) + 12 cos^{2} (θ) - 6

化简

12 - 18 cos (θ) + 12 cos^{2} (θ) - 6 : 12 cos^{2} (θ) - 18 cos (θ) + 6

12 - 18 cos (θ) + 12 cos^{2} (θ) - 6

对同类项分组

= - 18 cos (θ) + 12 cos^{2} (θ) + 12 - 6

数字相加/相减:

12 - 6 = 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

6 + 12 cos^{2} (θ) - 18 cos (θ) = 0

用替代法求解

6 + 12 cos^{2} (θ) - 18 cos (θ) = 0

令

: cos (θ) = u

6 + 12 u^{2} - 18 u = 0

6 + 12 u^{2} - 18 u = 0 : u = 1, u = \frac{1}{2}

6 + 12 u^{2} - 18 u = 0

改写成标准形式

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

12 u^{2} - 18 u + 6 = 0

使用求根公式求解

12 u^{2} - 18 u + 6 = 0

二次方程求根公式:

若

a = 12, b = - 18, c = 6

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

(- 18)^{2} - 4 \cdot 12 \cdot 6 = 6

(- 18)^{2} - 4 \cdot 12 \cdot 6

使用指数法则:

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

若

n

是偶数

(- 18)^{2} = 1 8^{2}

= 1 8^{2} - 4 \cdot 12 \cdot 6

数字相乘:

4 \cdot 12 \cdot 6 = 288

= 1 8^{2} - 288

1 8^{2} = 324

= 324 - 288

数字相减:

324 - 288 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 12}

将解分隔开

u_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 12}, u_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 12}

u = \frac{- ( - 18 ) + 6}{2 \cdot 12} : 1

\frac{- ( - 18 ) + 6}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{18 + 6}{2 \cdot 12}

数字相加:

18 + 6 = 24

= \frac{24}{2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{24}{24}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 18 ) - 6}{2 \cdot 12} : \frac{1}{2}

\frac{- ( - 18 ) - 6}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{18 - 6}{2 \cdot 12}

数字相减:

18 - 6 = 12

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

数字相乘:

2 \cdot 12 = 24

= \frac{12}{24}

约分:

12

= \frac{1}{2}

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1 : θ = 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 = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

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

合并所有解

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

作图

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