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受欢迎的三角函数 >

-5sin^2(θ)+17cos(θ)+9=4cos(θ)-2

解答

- 5 sin^{2} (θ) + 17 cos (θ) + 9 = 4 cos (θ) - 2

解答

θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

+1

度数

θ = 126.86989 \dots^{\circ} + 36 0^{\circ} n, θ = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 5 sin^{2} (θ) + 17 cos (θ) + 9 = 4 cos (θ) - 2

两边减去

4 cos (θ) - 2

13 cos (θ) - 5 sin^{2} (θ) + 11 = 0

使用三角恒等式改写

11 + 13 cos (θ) - 5 sin^{2} (θ)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= 11 + 13 cos (θ) - 5 (1 - cos^{2} (θ))

化简

11 + 13 cos (θ) - 5 (1 - cos^{2} (θ)) : 5 cos^{2} (θ) + 13 cos (θ) + 6

11 + 13 cos (θ) - 5 (1 - cos^{2} (θ))

乘开

- 5 (1 - cos^{2} (θ)) : - 5 + 5 cos^{2} (θ)

- 5 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 5, b = 1, c = cos^{2} (θ)

= - 5 \cdot 1 - (- 5) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 5 \cdot 1 + 5 cos^{2} (θ)

数字相乘:

5 \cdot 1 = 5

= - 5 + 5 cos^{2} (θ)

= 11 + 13 cos (θ) - 5 + 5 cos^{2} (θ)

化简

11 + 13 cos (θ) - 5 + 5 cos^{2} (θ) : 5 cos^{2} (θ) + 13 cos (θ) + 6

11 + 13 cos (θ) - 5 + 5 cos^{2} (θ)

对同类项分组

= 13 cos (θ) + 5 cos^{2} (θ) + 11 - 5

数字相加/相减:

11 - 5 = 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

6 + 13 cos (θ) + 5 cos^{2} (θ) = 0

用替代法求解

6 + 13 cos (θ) + 5 cos^{2} (θ) = 0

令

: cos (θ) = u

6 + 13 u + 5 u^{2} = 0

6 + 13 u + 5 u^{2} = 0 : u = - \frac{3}{5}, u = - 2

6 + 13 u + 5 u^{2} = 0

改写成标准形式

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

5 u^{2} + 13 u + 6 = 0

使用求根公式求解

5 u^{2} + 13 u + 6 = 0

二次方程求根公式:

若

a = 5, b = 13, c = 6

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

1 3^{2} - 4 \cdot 5 \cdot 6 = 7

1 3^{2} - 4 \cdot 5 \cdot 6

数字相乘:

4 \cdot 5 \cdot 6 = 120

= 1 3^{2} - 120

1 3^{2} = 169

= 169 - 120

数字相减:

169 - 120 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 13 \pm 7}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- 13 + 7}{2 \cdot 5}, u_{2} = \frac{- 13 - 7}{2 \cdot 5}

u = \frac{- 13 + 7}{2 \cdot 5} : - \frac{3}{5}

\frac{- 13 + 7}{2 \cdot 5}

数字相加/相减:

- 13 + 7 = - 6

= \frac{- 6}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 6}{10}

使用分式法则:

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

= - \frac{6}{10}

约分:

2

= - \frac{3}{5}

u = \frac{- 13 - 7}{2 \cdot 5} : - 2

\frac{- 13 - 7}{2 \cdot 5}

数字相减:

- 13 - 7 = - 20

= \frac{- 20}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 20}{10}

使用分式法则:

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

= - \frac{20}{10}

数字相除:

\frac{20}{10} = 2

= - 2

二次方程组的解是:

u = - \frac{3}{5}, u = - 2

u = cos (θ)

代回

cos (θ) = - \frac{3}{5}, cos (θ) = - 2

cos (θ) = - \frac{3}{5}, cos (θ) = - 2

cos (θ) = - \frac{3}{5} : θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

cos (θ) = - \frac{3}{5}

使用反三角函数性质

cos (θ) = - \frac{3}{5}

cos (θ) = - \frac{3}{5}

的通解

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

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

cos (θ) = - 2 :

无解

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

以小数形式表示解

θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

作图

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