zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

9sin^2(θ)+3cos(θ)-9=-2

解答

9 sin^{2} (θ) + 3 cos (θ) - 9 = - 2

解答

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

+1

度数

θ = 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = - 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

9 sin^{2} (θ) + 3 cos (θ) - 9 = - 2

两边减去

- 2

9 sin^{2} (θ) + 3 cos (θ) - 7 = 0

使用三角恒等式改写

- 7 + 3 cos (θ) + 9 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)

= - 7 + 3 cos (θ) + 9 (1 - cos^{2} (θ))

化简

- 7 + 3 cos (θ) + 9 (1 - cos^{2} (θ)) : 3 cos (θ) - 9 cos^{2} (θ) + 2

- 7 + 3 cos (θ) + 9 (1 - cos^{2} (θ))

乘开

9 (1 - cos^{2} (θ)) : 9 - 9 cos^{2} (θ)

9 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (θ)

= 9 \cdot 1 - 9 cos^{2} (θ)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (θ)

= - 7 + 3 cos (θ) + 9 - 9 cos^{2} (θ)

化简

- 7 + 3 cos (θ) + 9 - 9 cos^{2} (θ) : 3 cos (θ) - 9 cos^{2} (θ) + 2

- 7 + 3 cos (θ) + 9 - 9 cos^{2} (θ)

对同类项分组

= 3 cos (θ) - 9 cos^{2} (θ) - 7 + 9

数字相加/相减:

- 7 + 9 = 2

= 3 cos (θ) - 9 cos^{2} (θ) + 2

= 3 cos (θ) - 9 cos^{2} (θ) + 2

= 3 cos (θ) - 9 cos^{2} (θ) + 2

2 + 3 cos (θ) - 9 cos^{2} (θ) = 0

用替代法求解

2 + 3 cos (θ) - 9 cos^{2} (θ) = 0

令

: cos (θ) = u

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

2 + 3 u - 9 u^{2} = 0 : u = - \frac{1}{3}, u = \frac{2}{3}

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

改写成标准形式

a x^{2} + b x + c = 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 ( - 9 ) \cdot 2}{2 ( - 9 )}

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

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

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

使用法则

- (- a) = a

= 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 ( - 9 )}

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/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}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 3 - 9 = - 12

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

数字相乘:

2 \cdot 9 = 18

= \frac{- 12}{- 18}

使用分式法则:

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

= \frac{12}{18}

约分:

6

= \frac{2}{3}

二次方程组的解是:

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

u = cos (θ)

代回

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

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

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

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{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn, 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 = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn, θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

作图

流行的例子

2 cos (3 t) - 1 = 0

sin(x)+cos(x)-1=0

sin (x) + cos (x) - 1 = 0

tan (θ) = - 9

tan (θ) = 13

tan (θ) = 16