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2cos^4(x)+8sin^2(x)=5

解答

2 cos^{4} (x) + 8 sin^{2} (x) = 5

解答

x = 0.86689 \dots + 2 πn, x = 2 π - 0.86689 \dots + 2 πn, x = 2.27469 \dots + 2 πn, x = - 2.27469 \dots + 2 πn

+1

度数

x = 49.66956 \dots^{\circ} + 36 0^{\circ} n, x = 310.33043 \dots^{\circ} + 36 0^{\circ} n, x = 130.33043 \dots^{\circ} + 36 0^{\circ} n, x = - 130.33043 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{4} (x) + 8 sin^{2} (x) = 5

两边减去

5

2 cos^{4} (x) + 8 sin^{2} (x) - 5 = 0

使用三角恒等式改写

- 5 + 2 cos^{4} (x) + 8 sin^{2} (x)

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} (x) + 8 (1 - cos^{2} (x))

化简

- 5 + 2 cos^{4} (x) + 8 (1 - cos^{2} (x)) : 2 cos^{4} (x) - 8 cos^{2} (x) + 3

- 5 + 2 cos^{4} (x) + 8 (1 - cos^{2} (x))

乘开

8 (1 - cos^{2} (x)) : 8 - 8 cos^{2} (x)

8 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 8, b = 1, c = cos^{2} (x)

= 8 \cdot 1 - 8 cos^{2} (x)

数字相乘:

8 \cdot 1 = 8

= 8 - 8 cos^{2} (x)

= - 5 + 2 cos^{4} (x) + 8 - 8 cos^{2} (x)

化简

- 5 + 2 cos^{4} (x) + 8 - 8 cos^{2} (x) : 2 cos^{4} (x) - 8 cos^{2} (x) + 3

- 5 + 2 cos^{4} (x) + 8 - 8 cos^{2} (x)

对同类项分组

= 2 cos^{4} (x) - 8 cos^{2} (x) - 5 + 8

数字相加/相减:

- 5 + 8 = 3

= 2 cos^{4} (x) - 8 cos^{2} (x) + 3

= 2 cos^{4} (x) - 8 cos^{2} (x) + 3

= 2 cos^{4} (x) - 8 cos^{2} (x) + 3

3 + 2 cos^{4} (x) - 8 cos^{2} (x) = 0

用替代法求解

3 + 2 cos^{4} (x) - 8 cos^{2} (x) = 0

令

: cos (x) = u

3 + 2 u^{4} - 8 u^{2} = 0

3 + 2 u^{4} - 8 u^{2} = 0 : u = \frac{4 + 10}{2}, u = - \frac{4 + 10}{2}, u = \frac{4 - 10}{2}, u = - \frac{4 - 10}{2}

3 + 2 u^{4} - 8 u^{2} = 0

改写成标准形式

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

2 u^{4} - 8 u^{2} + 3 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - 8 v + 3 = 0

解

2 v^{2} - 8 v + 3 = 0 : v = \frac{4 + 10}{2}, v = \frac{4 - 10}{2}

2 v^{2} - 8 v + 3 = 0

使用求根公式求解

2 v^{2} - 8 v + 3 = 0

二次方程求根公式:

若

a = 2, b = - 8, c = 3

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 3 = 210

(- 8)^{2} - 4 \cdot 2 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

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

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 8^{2} - 24

8^{2} = 64

= 64 - 24

数字相减:

64 - 24 = 40

= 40

40

质因数分解:

2^{3} \cdot 5

40

40

除以

240 = 20 \cdot 2

= 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 5

使用根式运算法则:

= 2^{2} 2 \cdot 5

使用根式运算法则:

2^{2} = 2

= 2 2 \cdot 5

整理后得

= 210

v_{1, 2} = \frac{- ( - 8 ) \pm 2 10}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 8 ) + 2 10}{2 \cdot 2}, v_{2} = \frac{- ( - 8 ) - 2 10}{2 \cdot 2}

v = \frac{- ( - 8 ) + 2 10}{2 \cdot 2} : \frac{4 + 10}{2}

\frac{- ( - 8 ) + 2 10}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{8 + 2 10}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8 + 2 10}{4}

分解

8 + 210 : 2 (4 + 10)

8 + 210

改写为

= 2 \cdot 4 + 210

因式分解出通项

2

= 2 (4 + 10)

= \frac{2 ( 4 + 10 )}{4}

约分:

2

= \frac{4 + 10}{2}

v = \frac{- ( - 8 ) - 2 10}{2 \cdot 2} : \frac{4 - 10}{2}

\frac{- ( - 8 ) - 2 10}{2 \cdot 2}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{8 - 2 10}{4}

分解

8 - 210 : 2 (4 - 10)

8 - 210

改写为

= 2 \cdot 4 - 210

因式分解出通项

2

= 2 (4 - 10)

= \frac{2 ( 4 - 10 )}{4}

约分:

2

= \frac{4 - 10}{2}

二次方程组的解是:

v = \frac{4 + 10}{2}, v = \frac{4 - 10}{2}

v = \frac{4 + 10}{2}, v = \frac{4 - 10}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = \frac{4 + 10}{2} : u = \frac{4 + 10}{2}, u = - \frac{4 + 10}{2}

u^{2} = \frac{4 + 10}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{4 + 10}{2}, u = - \frac{4 + 10}{2}

解

u^{2} = \frac{4 - 10}{2} : u = \frac{4 - 10}{2}, u = - \frac{4 - 10}{2}

u^{2} = \frac{4 - 10}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{4 - 10}{2}, u = - \frac{4 - 10}{2}

解为

u = \frac{4 + 10}{2}, u = - \frac{4 + 10}{2}, u = \frac{4 - 10}{2}, u = - \frac{4 - 10}{2}

u = cos (x)

代回

cos (x) = \frac{4 + 10}{2}, cos (x) = - \frac{4 + 10}{2}, cos (x) = \frac{4 - 10}{2}, cos (x) = - \frac{4 - 10}{2}

cos (x) = \frac{4 + 10}{2}, cos (x) = - \frac{4 + 10}{2}, cos (x) = \frac{4 - 10}{2}, cos (x) = - \frac{4 - 10}{2}

cos (x) = \frac{4 + 10}{2} :

无解

cos (x) = \frac{4 + 10}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = - \frac{4 + 10}{2} :

无解

cos (x) = - \frac{4 + 10}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{4 - 10}{2} : x = arccos \frac{4 - 10}{2} + 2 πn, x = 2 π - arccos \frac{4 - 10}{2} + 2 πn

cos (x) = \frac{4 - 10}{2}

使用反三角函数性质

cos (x) = \frac{4 - 10}{2}

cos (x) = \frac{4 - 10}{2}

的通解

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

x = arccos \frac{4 - 10}{2} + 2 πn, x = 2 π - arccos \frac{4 - 10}{2} + 2 πn

x = arccos \frac{4 - 10}{2} + 2 πn, x = 2 π - arccos \frac{4 - 10}{2} + 2 πn

cos (x) = - \frac{4 - 10}{2} : x = arccos - \frac{4 - 10}{2} + 2 πn, x = - arccos - \frac{4 - 10}{2} + 2 πn

cos (x) = - \frac{4 - 10}{2}

使用反三角函数性质

cos (x) = - \frac{4 - 10}{2}

cos (x) = - \frac{4 - 10}{2}

的通解

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

x = arccos - \frac{4 - 10}{2} + 2 πn, x = - arccos - \frac{4 - 10}{2} + 2 πn

x = arccos - \frac{4 - 10}{2} + 2 πn, x = - arccos - \frac{4 - 10}{2} + 2 πn

合并所有解

x = arccos \frac{4 - 10}{2} + 2 πn, x = 2 π - arccos \frac{4 - 10}{2} + 2 πn, x = arccos - \frac{4 - 10}{2} + 2 πn, x = - arccos - \frac{4 - 10}{2} + 2 πn

以小数形式表示解

x = 0.86689 \dots + 2 πn, x = 2 π - 0.86689 \dots + 2 πn, x = 2.27469 \dots + 2 πn, x = - 2.27469 \dots + 2 πn

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