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受欢迎的三角函数 >

9cos^2(θ)-10cos(θ)-2=-7cos(θ)

解答

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

解答

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

+1

度数

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

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

用替代法求解

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

令

: cos (θ) = u

9 u^{2} - 10 u - 2 = - 7 u

9 u^{2} - 10 u - 2 = - 7 u : u = \frac{2}{3}, u = - \frac{1}{3}

9 u^{2} - 10 u - 2 = - 7 u

将

7 u

para o lado esquerdo

9 u^{2} - 10 u - 2 = - 7 u

两边加上

7 u

9 u^{2} - 10 u - 2 + 7 u = - 7 u + 7 u

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 9, b = - 3, c = - 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 9 \cdot 2

数字相乘:

4 \cdot 9 \cdot 2 = 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 9}

将解分隔开

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

u = \frac{- ( - 3 ) + 9}{2 \cdot 9} : \frac{2}{3}

\frac{- ( - 3 ) + 9}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{3 + 9}{2 \cdot 9}

数字相加:

3 + 9 = 12

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

数字相乘:

2 \cdot 9 = 18

= \frac{12}{18}

约分:

6

= \frac{2}{3}

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

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

使用法则

- (- a) = a

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

数字相减:

3 - 9 = - 6

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

数字相乘:

2 \cdot 9 = 18

= \frac{- 6}{18}

使用分式法则:

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

= - \frac{6}{18}

约分:

6

= - \frac{1}{3}

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = \frac{2}{3}, cos (θ) = - \frac{1}{3}

cos (θ) = \frac{2}{3}, cos (θ) = - \frac{1}{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

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 = arccos (\frac{2}{3}) + 2 πn, θ = 2 π - arccos (\frac{2}{3}) + 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 = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

作图

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