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受欢迎的三角函数 >

3cos(2θ)=-3cos(θ)

解答

3 cos (2 θ) = - 3 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

3 cos (2 θ) = - 3 cos (θ)

两边减去

- 3 cos (θ)

3 cos (2 θ) + 3 cos (θ) = 0

使用三角恒等式改写

3 cos (2 θ) + 3 cos (θ)

使用倍角公式:

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

= 3 (2 cos^{2} (θ) - 1) + 3 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 3 + 3 cos (θ) = 0

用替代法求解

(- 1 + 2 cos^{2} (θ)) \cdot 3 + 3 cos (θ) = 0

令

: cos (θ) = u

(- 1 + 2 u^{2}) \cdot 3 + 3 u = 0

(- 1 + 2 u^{2}) \cdot 3 + 3 u = 0 : u = \frac{1}{2}, u = - 1

(- 1 + 2 u^{2}) \cdot 3 + 3 u = 0

展开

(- 1 + 2 u^{2}) \cdot 3 + 3 u : - 3 + 6 u^{2} + 3 u

(- 1 + 2 u^{2}) \cdot 3 + 3 u

= 3 (- 1 + 2 u^{2}) + 3 u

乘开

3 (- 1 + 2 u^{2}) : - 3 + 6 u^{2}

3 (- 1 + 2 u^{2})

使用分配律:

a (b + c) = ab + a c

a = 3, b = - 1, c = 2 u^{2}

= 3 (- 1) + 3 \cdot 2 u^{2}

使用加减运算法则

+ (- a) = - a

= - 3 \cdot 1 + 3 \cdot 2 u^{2}

化简

- 3 \cdot 1 + 3 \cdot 2 u^{2} : - 3 + 6 u^{2}

- 3 \cdot 1 + 3 \cdot 2 u^{2}

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 u^{2}

数字相乘:

3 \cdot 2 = 6

= - 3 + 6 u^{2}

= - 3 + 6 u^{2}

= - 3 + 6 u^{2} + 3 u

- 3 + 6 u^{2} + 3 u = 0

改写成标准形式

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

6 u^{2} + 3 u - 3 = 0

使用求根公式求解

6 u^{2} + 3 u - 3 = 0

二次方程求根公式:

若

a = 6, b = 3, c = - 3

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

3^{2} - 4 \cdot 6 (- 3) = 9

3^{2} - 4 \cdot 6 (- 3)

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 6 \cdot 3

数字相乘:

4 \cdot 6 \cdot 3 = 72

= 3^{2} + 72

3^{2} = 9

= 9 + 72

数字相加:

9 + 72 = 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{- 3 \pm 9}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 3 + 9}{2 \cdot 6}, u_{2} = \frac{- 3 - 9}{2 \cdot 6}

u = \frac{- 3 + 9}{2 \cdot 6} : \frac{1}{2}

\frac{- 3 + 9}{2 \cdot 6}

数字相加/相减:

- 3 + 9 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u = \frac{- 3 - 9}{2 \cdot 6} : - 1

\frac{- 3 - 9}{2 \cdot 6}

数字相减:

- 3 - 9 = - 12

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 12}{12}

使用分式法则:

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

= - \frac{12}{12}

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

流行的例子

csc^2(θ)+3csc(θ)+2=0

csc^{2} (θ) + 3 csc (θ) + 2 = 0

cos^2(x)=3-5sin(x)

cos^{2} (x) = 3 - 5 sin (x)

8sin(x/2)+8cos(x)=0

8 sin (\frac{x}{2}) + 8 cos (x) = 0

tan(θ/2+pi/6)=1

tan (\frac{θ}{2} + \frac{π}{6}) = 1

sec(x)=-sqrt(2),0<x<2pi

sec (x) = - 2, 0 < x < 2 π