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受欢迎的三角函数 >

3cos^2(θ)-1=2cos(θ)

解答

3 cos^{2} (θ) - 1 = 2 cos (θ)

解答

θ = 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos^{2} (θ) - 1 = 2 cos (θ)

用替代法求解

3 cos^{2} (θ) - 1 = 2 cos (θ)

令

: cos (θ) = u

3 u^{2} - 1 = 2 u

3 u^{2} - 1 = 2 u : u = 1, u = - \frac{1}{3}

3 u^{2} - 1 = 2 u

将

2 u

para o lado esquerdo

3 u^{2} - 1 = 2 u

两边减去

2 u

3 u^{2} - 1 - 2 u = 2 u - 2 u

化简

3 u^{2} - 1 - 2 u = 0

3 u^{2} - 1 - 2 u = 0

改写成标准形式

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

3 u^{2} - 2 u - 1 = 0

使用求根公式求解

3 u^{2} - 2 u - 1 = 0

二次方程求根公式:

若

a = 3, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 3 (- 1) = 4

(- 2)^{2} - 4 \cdot 3 (- 1)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 3 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 3 \cdot 1

数字相乘:

4 \cdot 3 \cdot 1 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数字相加:

4 + 12 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

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

将解分隔开

u_{1} = \frac{- ( - 2 ) + 4}{2 \cdot 3}, u_{2} = \frac{- ( - 2 ) - 4}{2 \cdot 3}

u = \frac{- ( - 2 ) + 4}{2 \cdot 3} : 1

\frac{- ( - 2 ) + 4}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{2 + 4}{2 \cdot 3}

数字相加:

2 + 4 = 6

= \frac{6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{6}{6}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 4}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 2 ) - 4}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{2 - 4}{2 \cdot 3}

数字相减:

2 - 4 = - 2

= \frac{- 2}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 2}{6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = 1, cos (θ) = - \frac{1}{3}

cos (θ) = 1, cos (θ) = - \frac{1}{3}

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}{3} : θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3}

使用反三角函数性质

cos (θ) = - \frac{1}{3}

cos (θ) = - \frac{1}{3}

的通解

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

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

合并所有解

θ = 2 πn, θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

以小数形式表示解

θ = 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

作图

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