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受欢迎的三角函数 >

9cos(2θ)=-9cos(θ)

解答

9 cos (2 θ) = - 9 cos (θ)

解答

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

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

9 cos (2 θ) = - 9 cos (θ)

两边减去

- 9 cos (θ)

9 cos (2 θ) + 9 cos (θ) = 0

使用三角恒等式改写

9 cos (2 θ) + 9 cos (θ)

使用倍角公式:

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

= 9 (2 cos^{2} (θ) - 1) + 9 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 9 + 9 cos (θ) = 0

用替代法求解

(- 1 + 2 cos^{2} (θ)) \cdot 9 + 9 cos (θ) = 0

令

: cos (θ) = u

(- 1 + 2 u^{2}) \cdot 9 + 9 u = 0

(- 1 + 2 u^{2}) \cdot 9 + 9 u = 0 : u = \frac{1}{2}, u = - 1

(- 1 + 2 u^{2}) \cdot 9 + 9 u = 0

展开

(- 1 + 2 u^{2}) \cdot 9 + 9 u : - 9 + 18 u^{2} + 9 u

(- 1 + 2 u^{2}) \cdot 9 + 9 u

= 9 (- 1 + 2 u^{2}) + 9 u

乘开

9 (- 1 + 2 u^{2}) : - 9 + 18 u^{2}

9 (- 1 + 2 u^{2})

使用分配律:

a (b + c) = ab + a c

a = 9, b = - 1, c = 2 u^{2}

= 9 (- 1) + 9 \cdot 2 u^{2}

使用加减运算法则

+ (- a) = - a

= - 9 \cdot 1 + 9 \cdot 2 u^{2}

化简

- 9 \cdot 1 + 9 \cdot 2 u^{2} : - 9 + 18 u^{2}

- 9 \cdot 1 + 9 \cdot 2 u^{2}

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 \cdot 2 u^{2}

数字相乘:

9 \cdot 2 = 18

= - 9 + 18 u^{2}

= - 9 + 18 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

= 9^{2} + 4 \cdot 18 \cdot 9

数字相乘:

4 \cdot 18 \cdot 9 = 648

= 9^{2} + 648

9^{2} = 81

= 81 + 648

数字相加:

81 + 648 = 729

= 729

因式分解数字:

729 = 2 7^{2}

= 2 7^{2}

使用根式运算法则:

n a^{n} = a

2 7^{2} = 27

= 27

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

将解分隔开

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

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

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

数字相加/相减:

- 9 + 27 = 18

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

数字相乘:

2 \cdot 18 = 36

= \frac{18}{36}

约分:

18

= \frac{1}{2}

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

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

数字相减:

- 9 - 27 = - 36

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

数字相乘:

2 \cdot 18 = 36

= \frac{- 36}{36}

使用分式法则:

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

= - \frac{36}{36}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2}, cos (θ) = - 1

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

作图

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