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受欢迎的三角函数 >

9cos(2θ)=27cos(θ)-18

解答

9 cos (2 θ) = 27 cos (θ) - 18

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

9 cos (2 θ) = 27 cos (θ) - 18

两边减去

27 cos (θ) - 18

9 cos (2 θ) - 27 cos (θ) + 18 = 0

使用三角恒等式改写

18 - 27 cos (θ) + 9 cos (2 θ)

使用倍角公式:

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

= 18 - 27 cos (θ) + 9 (2 cos^{2} (θ) - 1)

化简

18 - 27 cos (θ) + 9 (2 cos^{2} (θ) - 1) : 18 cos^{2} (θ) - 27 cos (θ) + 9

18 - 27 cos (θ) + 9 (2 cos^{2} (θ) - 1)

乘开

9 (2 cos^{2} (θ) - 1) : 18 cos^{2} (θ) - 9

9 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 9, b = 2 cos^{2} (θ), c = 1

= 9 \cdot 2 cos^{2} (θ) - 9 \cdot 1

化简

9 \cdot 2 cos^{2} (θ) - 9 \cdot 1 : 18 cos^{2} (θ) - 9

9 \cdot 2 cos^{2} (θ) - 9 \cdot 1

数字相乘:

9 \cdot 2 = 18

= 18 cos^{2} (θ) - 9 \cdot 1

数字相乘:

9 \cdot 1 = 9

= 18 cos^{2} (θ) - 9

= 18 cos^{2} (θ) - 9

= 18 - 27 cos (θ) + 18 cos^{2} (θ) - 9

化简

18 - 27 cos (θ) + 18 cos^{2} (θ) - 9 : 18 cos^{2} (θ) - 27 cos (θ) + 9

18 - 27 cos (θ) + 18 cos^{2} (θ) - 9

对同类项分组

= - 27 cos (θ) + 18 cos^{2} (θ) + 18 - 9

数字相加/相减:

18 - 9 = 9

= 18 cos^{2} (θ) - 27 cos (θ) + 9

= 18 cos^{2} (θ) - 27 cos (θ) + 9

= 18 cos^{2} (θ) - 27 cos (θ) + 9

9 + 18 cos^{2} (θ) - 27 cos (θ) = 0

用替代法求解

9 + 18 cos^{2} (θ) - 27 cos (θ) = 0

令

: cos (θ) = u

9 + 18 u^{2} - 27 u = 0

9 + 18 u^{2} - 27 u = 0 : u = 1, u = \frac{1}{2}

9 + 18 u^{2} - 27 u = 0

改写成标准形式

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

18 u^{2} - 27 u + 9 = 0

使用求根公式求解

18 u^{2} - 27 u + 9 = 0

二次方程求根公式:

若

a = 18, b = - 27, c = 9

u_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 18 \cdot 9}{2 \cdot 18}

u_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 18 \cdot 9}{2 \cdot 18}

(- 27)^{2} - 4 \cdot 18 \cdot 9 = 9

(- 27)^{2} - 4 \cdot 18 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 27)^{2} = 2 7^{2}

= 2 7^{2} - 4 \cdot 18 \cdot 9

数字相乘:

4 \cdot 18 \cdot 9 = 648

= 2 7^{2} - 648

2 7^{2} = 729

= 729 - 648

数字相减:

729 - 648 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 27 ) \pm 9}{2 \cdot 18}

将解分隔开

u_{1} = \frac{- ( - 27 ) + 9}{2 \cdot 18}, u_{2} = \frac{- ( - 27 ) - 9}{2 \cdot 18}

u = \frac{- ( - 27 ) + 9}{2 \cdot 18} : 1

\frac{- ( - 27 ) + 9}{2 \cdot 18}

使用法则

- (- a) = a

= \frac{27 + 9}{2 \cdot 18}

数字相加:

27 + 9 = 36

= \frac{36}{2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{36}{36}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 27 ) - 9}{2 \cdot 18} : \frac{1}{2}

\frac{- ( - 27 ) - 9}{2 \cdot 18}

使用法则

- (- a) = a

= \frac{27 - 9}{2 \cdot 18}

数字相减:

27 - 9 = 18

= \frac{18}{2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{18}{36}

约分:

18

= \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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