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受欢迎的三角函数 >

5sin^2(θ)-4cos(θ)-9=-cos(θ)-6

解答

5 sin^{2} (θ) - 4 cos (θ) - 9 = - cos (θ) - 6

解答

θ = π + 2 πn, θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

+1

度数

θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 66.42182 \dots^{\circ} + 36 0^{\circ} n, θ = 293.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 sin^{2} (θ) - 4 cos (θ) - 9 = - cos (θ) - 6

两边减去

- cos (θ) - 6

5 sin^{2} (θ) - 3 cos (θ) - 3 = 0

使用三角恒等式改写

- 3 - 3 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)

= - 3 - 3 cos (θ) + 5 (1 - cos^{2} (θ))

化简

- 3 - 3 cos (θ) + 5 (1 - cos^{2} (θ)) : - 5 cos^{2} (θ) - 3 cos (θ) + 2

- 3 - 3 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)

数字相乘:

5 \cdot 1 = 5

= 5 - 5 cos^{2} (θ)

= - 3 - 3 cos (θ) + 5 - 5 cos^{2} (θ)

化简

- 3 - 3 cos (θ) + 5 - 5 cos^{2} (θ) : - 5 cos^{2} (θ) - 3 cos (θ) + 2

- 3 - 3 cos (θ) + 5 - 5 cos^{2} (θ)

对同类项分组

= - 3 cos (θ) - 5 cos^{2} (θ) - 3 + 5

数字相加/相减:

- 3 + 5 = 2

= - 5 cos^{2} (θ) - 3 cos (θ) + 2

= - 5 cos^{2} (θ) - 3 cos (θ) + 2

= - 5 cos^{2} (θ) - 3 cos (θ) + 2

2 - 3 cos (θ) - 5 cos^{2} (θ) = 0

用替代法求解

2 - 3 cos (θ) - 5 cos^{2} (θ) = 0

令

: cos (θ) = u

2 - 3 u - 5 u^{2} = 0

2 - 3 u - 5 u^{2} = 0 : u = - 1, u = \frac{2}{5}

2 - 3 u - 5 u^{2} = 0

改写成标准形式

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

- 5 u^{2} - 3 u + 2 = 0

使用求根公式求解

- 5 u^{2} - 3 u + 2 = 0

二次方程求根公式:

若

a = - 5, b = - 3, c = 2

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

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

(- 3)^{2} - 4 (- 5) \cdot 2 = 7

(- 3)^{2} - 4 (- 5) \cdot 2

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 5 \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 5 \cdot 2

数字相乘:

4 \cdot 5 \cdot 2 = 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

数字相加:

9 + 40 = 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{- ( - 3 ) \pm 7}{2 ( - 5 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 7}{2 ( - 5 )}, u_{2} = \frac{- ( - 3 ) - 7}{2 ( - 5 )}

u = \frac{- ( - 3 ) + 7}{2 ( - 5 )} : - 1

\frac{- ( - 3 ) + 7}{2 ( - 5 )}

去除括号:

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

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

数字相加:

3 + 7 = 10

= \frac{10}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10}{- 10}

使用分式法则:

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

= - \frac{10}{10}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 3 ) - 7}{2 ( - 5 )} : \frac{2}{5}

\frac{- ( - 3 ) - 7}{2 ( - 5 )}

去除括号:

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

= \frac{3 - 7}{- 2 \cdot 5}

数字相减:

3 - 7 = - 4

= \frac{- 4}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 4}{- 10}

使用分式法则:

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

= \frac{4}{10}

约分:

2

= \frac{2}{5}

二次方程组的解是:

u = - 1, u = \frac{2}{5}

u = cos (θ)

代回

cos (θ) = - 1, cos (θ) = \frac{2}{5}

cos (θ) = - 1, cos (θ) = \frac{2}{5}

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

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

cos (θ) = \frac{2}{5}

使用反三角函数性质

cos (θ) = \frac{2}{5}

cos (θ) = \frac{2}{5}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = π + 2 πn, θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

作图

流行的例子

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12 sin^{2} (θ) - 6 sin (θ) = 5

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cos (c) = 0.87

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