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受欢迎的三角函数 >

-3sin^2(θ)+3=-10cos(θ)-3

解答

- 3 sin^{2} (θ) + 3 = - 10 cos (θ) - 3

解答

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

+1

度数

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

求解步骤

- 3 sin^{2} (θ) + 3 = - 10 cos (θ) - 3

两边减去

- 10 cos (θ) - 3

10 cos (θ) - 3 sin^{2} (θ) + 6 = 0

使用三角恒等式改写

6 + 10 cos (θ) - 3 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)

= 6 + 10 cos (θ) - 3 (1 - cos^{2} (θ))

化简

6 + 10 cos (θ) - 3 (1 - cos^{2} (θ)) : 3 cos^{2} (θ) + 10 cos (θ) + 3

6 + 10 cos (θ) - 3 (1 - cos^{2} (θ))

乘开

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

- 3 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (θ)

= - 3 \cdot 1 - (- 3) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 cos^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 cos^{2} (θ)

= 6 + 10 cos (θ) - 3 + 3 cos^{2} (θ)

化简

6 + 10 cos (θ) - 3 + 3 cos^{2} (θ) : 3 cos^{2} (θ) + 10 cos (θ) + 3

6 + 10 cos (θ) - 3 + 3 cos^{2} (θ)

对同类项分组

= 10 cos (θ) + 3 cos^{2} (θ) + 6 - 3

数字相加/相减:

6 - 3 = 3

= 3 cos^{2} (θ) + 10 cos (θ) + 3

= 3 cos^{2} (θ) + 10 cos (θ) + 3

= 3 cos^{2} (θ) + 10 cos (θ) + 3

3 + 10 cos (θ) + 3 cos^{2} (θ) = 0

用替代法求解

3 + 10 cos (θ) + 3 cos^{2} (θ) = 0

令

: cos (θ) = u

3 + 10 u + 3 u^{2} = 0

3 + 10 u + 3 u^{2} = 0 : u = - \frac{1}{3}, u = - 3

3 + 10 u + 3 u^{2} = 0

改写成标准形式

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

3 u^{2} + 10 u + 3 = 0

使用求根公式求解

3 u^{2} + 10 u + 3 = 0

二次方程求根公式:

若

a = 3, b = 10, c = 3

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 3 = 8

1 0^{2} - 4 \cdot 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

数字相减:

100 - 36 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- 10 \pm 8}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 10 + 8}{2 \cdot 3}, u_{2} = \frac{- 10 - 8}{2 \cdot 3}

u = \frac{- 10 + 8}{2 \cdot 3} : - \frac{1}{3}

\frac{- 10 + 8}{2 \cdot 3}

数字相加/相减:

- 10 + 8 = - 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{- 2}{6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

u = \frac{- 10 - 8}{2 \cdot 3} : - 3

\frac{- 10 - 8}{2 \cdot 3}

数字相减:

- 10 - 8 = - 18

= \frac{- 18}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 18}{6}

使用分式法则:

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

= - \frac{18}{6}

数字相除:

\frac{18}{6} = 3

= - 3

二次方程组的解是:

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

u = cos (θ)

代回

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

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

无解

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

以小数形式表示解

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

作图

流行的例子

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cot(3x)=(sqrt(3))/3

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