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受欢迎的三角函数 >

9cos^2(θ)-12cos(θ)=-4

解答

9 cos^{2} (θ) - 12 cos (θ) = - 4

解答

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

+1

度数

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

9 cos^{2} (θ) - 12 cos (θ) = - 4

用替代法求解

9 cos^{2} (θ) - 12 cos (θ) = - 4

令

: cos (θ) = u

9 u^{2} - 12 u = - 4

9 u^{2} - 12 u = - 4 : u = \frac{2}{3}

9 u^{2} - 12 u = - 4

将

4

para o lado esquerdo

9 u^{2} - 12 u = - 4

两边加上

4

9 u^{2} - 12 u + 4 = - 4 + 4

化简

9 u^{2} - 12 u + 4 = 0

9 u^{2} - 12 u + 4 = 0

使用求根公式求解

9 u^{2} - 12 u + 4 = 0

二次方程求根公式:

若

a = 9, b = - 12, c = 4

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

(- 12)^{2} - 4 \cdot 9 \cdot 4 = 0

(- 12)^{2} - 4 \cdot 9 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 9 \cdot 4

数字相乘:

4 \cdot 9 \cdot 4 = 144

= 1 2^{2} - 144

1 2^{2} = 144

= 144 - 144

数字相减:

144 - 144 = 0

= 0

u_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 9}

u = \frac{- ( - 12 )}{2 \cdot 9}

\frac{- ( - 12 )}{2 \cdot 9} = \frac{2}{3}

\frac{- ( - 12 )}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{12}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{12}{18}

约分:

6

= \frac{2}{3}

u = \frac{2}{3}

二次方程组的解是:

u = \frac{2}{3}

u = cos (θ)

代回

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

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

cos (θ) = \frac{2}{3}

使用反三角函数性质

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

作图

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