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受欢迎的三角函数 >

2cos^4(θ)+9sin^2(θ)=5

解答

2 cos^{4} (θ) + 9 sin^{2} (θ) = 5

解答

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

+1

度数

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{4} (θ) + 9 sin^{2} (θ) = 5

两边减去

5

2 cos^{4} (θ) + 9 sin^{2} (θ) - 5 = 0

使用三角恒等式改写

- 5 + 2 cos^{4} (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)

= - 5 + 2 cos^{4} (θ) + 9 (1 - cos^{2} (θ))

化简

- 5 + 2 cos^{4} (θ) + 9 (1 - cos^{2} (θ)) : 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

- 5 + 2 cos^{4} (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)

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

化简

- 5 + 2 cos^{4} (θ) + 9 - 9 cos^{2} (θ) : 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

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

对同类项分组

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

数字相加/相减:

- 5 + 9 = 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

4 + 2 cos^{4} (θ) - 9 cos^{2} (θ) = 0

用替代法求解

4 + 2 cos^{4} (θ) - 9 cos^{2} (θ) = 0

令

: cos (θ) = u

4 + 2 u^{4} - 9 u^{2} = 0

4 + 2 u^{4} - 9 u^{2} = 0 : u = 2, u = - 2, u = \frac{1}{2}, u = - \frac{1}{2}

4 + 2 u^{4} - 9 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 9 u^{2} + 4 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - 9 v + 4 = 0

解

2 v^{2} - 9 v + 4 = 0 : v = 4, v = \frac{1}{2}

2 v^{2} - 9 v + 4 = 0

使用求根公式求解

2 v^{2} - 9 v + 4 = 0

二次方程求根公式:

若

a = 2, b = - 9, c = 4

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

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

(- 9)^{2} - 4 \cdot 2 \cdot 4 = 7

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

使用指数法则:

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

若

n

是偶数

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 2 \cdot 4

数字相乘:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数字相减:

81 - 32 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

7^{2} = 7

= 7

v_{1, 2} = \frac{- ( - 9 ) \pm 7}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 9 ) + 7}{2 \cdot 2}, v_{2} = \frac{- ( - 9 ) - 7}{2 \cdot 2}

v = \frac{- ( - 9 ) + 7}{2 \cdot 2} : 4

\frac{- ( - 9 ) + 7}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{9 + 7}{2 \cdot 2}

数字相加:

9 + 7 = 16

= \frac{16}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{16}{4}

数字相除:

\frac{16}{4} = 4

= 4

v = \frac{- ( - 9 ) - 7}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 9 ) - 7}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{9 - 7}{2 \cdot 2}

数字相减:

9 - 7 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

v = 4, v = \frac{1}{2}

v = 4, v = \frac{1}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 4, u = - 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

解

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

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

解为

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

u = cos (θ)

代回

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

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

cos (θ) = 2 :

无解

cos (θ) = 2

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - 2 :

无解

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

无解

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

cos (θ) = \frac{1}{2}

使用反三角函数性质

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

的通解

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

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

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

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

cos (θ) = - \frac{1}{2}

使用反三角函数性质

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

作图

流行的例子

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