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受欢迎的三角函数 >

-6cos^2(θ)-9cos(θ)-7=8cos(θ)

解答

- 6 cos^{2} (θ) - 9 cos (θ) - 7 = 8 cos (θ)

解答

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

- 6 cos^{2} (θ) - 9 cos (θ) - 7 = 8 cos (θ)

用替代法求解

- 6 cos^{2} (θ) - 9 cos (θ) - 7 = 8 cos (θ)

令

: cos (θ) = u

- 6 u^{2} - 9 u - 7 = 8 u

- 6 u^{2} - 9 u - 7 = 8 u : u = - \frac{7}{3}, u = - \frac{1}{2}

- 6 u^{2} - 9 u - 7 = 8 u

将

8 u

para o lado esquerdo

- 6 u^{2} - 9 u - 7 = 8 u

两边减去

8 u

- 6 u^{2} - 9 u - 7 - 8 u = 8 u - 8 u

化简

- 6 u^{2} - 17 u - 7 = 0

- 6 u^{2} - 17 u - 7 = 0

使用求根公式求解

- 6 u^{2} - 17 u - 7 = 0

二次方程求根公式:

若

a = - 6, b = - 17, c = - 7

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 ( - 6 ) ( - 7 )}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 ( - 6 ) ( - 7 )}{2 ( - 6 )}

(- 17)^{2} - 4 (- 6) (- 7) = 11

(- 17)^{2} - 4 (- 6) (- 7)

使用法则

- (- a) = a

= (- 17)^{2} - 4 \cdot 6 \cdot 7

使用指数法则:

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

若

n

是偶数

(- 17)^{2} = 1 7^{2}

= 1 7^{2} - 4 \cdot 6 \cdot 7

数字相乘:

4 \cdot 6 \cdot 7 = 168

= 1 7^{2} - 168

1 7^{2} = 289

= 289 - 168

数字相减:

289 - 168 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 17 ) \pm 11}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- ( - 17 ) + 11}{2 ( - 6 )}, u_{2} = \frac{- ( - 17 ) - 11}{2 ( - 6 )}

u = \frac{- ( - 17 ) + 11}{2 ( - 6 )} : - \frac{7}{3}

\frac{- ( - 17 ) + 11}{2 ( - 6 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{17 + 11}{- 2 \cdot 6}

数字相加:

17 + 11 = 28

= \frac{28}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{28}{- 12}

使用分式法则:

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

= - \frac{28}{12}

约分:

4

= - \frac{7}{3}

u = \frac{- ( - 17 ) - 11}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 17 ) - 11}{2 ( - 6 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{17 - 11}{- 2 \cdot 6}

数字相减:

17 - 11 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

二次方程组的解是:

u = - \frac{7}{3}, u = - \frac{1}{2}

u = cos (θ)

代回

cos (θ) = - \frac{7}{3}, cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{7}{3}, cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{7}{3} :

无解

cos (θ) = - \frac{7}{3}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

合并所有解

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

作图

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